Online Instructor’s Manual
to accompany
Introductory Circuit Analysis
Twelfth Edition
Robert L. Boylestad
Prentice Hall
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Contents
CHAPTER 1
1
CHAPTER 2
10
CHAPTER 3
15
CHAPTER 4
25
CHAPTER 5
32
CHAPTER 6
42
CHAPTER 7
56
CHAPTER 8
68
CHAPTER 9
92
CHAPTER 10
111
CHAPTER 11
130
CHAPTER 12
149
CHAPTER 13
156
CHAPTER 14
164
CHAPTER 15
174
CHAPTER 16
196
CHAPTER 17
203
CHAPTER 18
222
CHAPTER 19
252
CHAPTER 20
265
CHAPTER 21
279
CHAPTER 22
311
CHAPTER 23
318
CHAPTER 24
333
CHAPTER 25
342
CHAPTER 25
352
iii
___________________________________________________________________________________________
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10 9 8 7 6 5 4 3 2 1
ISBN-13: 978-0-13-504082-9
ISBN-10:
0-13-504082-5
Chapter 1
1.

2.

3.

4.
=
5.
6.
d 20,000 ft  1 mi   60 s   60 min 
=
= 1363.64 mph
t
10 s  5,280 ft  1 min   1 h 
 1h 
4 min 
 = 0.067 h
 60 min 
d
31 mi
= 
= 29.05 mph
t 1.067 h
 3 ft   1 mi 
100 yds 

 = 0.0568 mi
1 yd   5,280 ft 
60 mi  1 h  1 min 
= 0.0167 mi/s
h  60 min   60 s 
t=
7.
a.
b.
c.
d


0.0568 mi
= 3.40 s
0.0167 mi/s
95 mi  5,280 ft   1 h  1 min 
= 139.33 ft/s
h  mi   60 min   60 s 
d
60 ft
t= 
= 0.431 s
 139.33 ft/s
d 60 ft  60 s   60 min   1 mi 
= 
= 40.91 mph
t
1 s 1 min   1 h   5,280 ft 
8.

9.

10.

11.
MKS, CGS, C =
12.
 0.7378 ft - lb 
1000 J 
 = 737.8 ft-lbs
1J


5
5
5
(F  32)  (68  32)  (36) = 20
9
9
9
SI: K = 273.15 + C = 273.15 + 20 = 293.15
Chapter 1
1
13.
 3 ft  12 in.   2.54 cm 
0.5 yd 


 = 45.72 cm
1 yd   1 ft   1 in. 
14.
a.
°F = 2(°C) + 30° = 40° + 30° = 70°
b.
°F =
c.
very close
d.
30°C  90°F vs. 86°
5°C  40°F vs 41°
a.
14.6
b.
c.
1,046.1
d.
e.
3.14159 = 3.1
a.
14.60
b.
c.
1,046.06
d.
e.
3.14159 = 3.14
a.
14.603
b.
c.
1,046.060
d.
e.
3.14159 = 3.142
a.
104
15.
16.
17.
18.
9
9
(C)  32  (20)  32 = 68°
5
5
e. 100
b. 106
56.0
1
= 0.0625 = 0.1
16
56.04
1
= 0.0625 = 0.06
16
56.042
1
= 0.0625 = 0.063
16
c. 103
d. 103
c. 2.4  106
d. 60  103
f. 101
5  103
a.
15  103
e.
4.02  104 f. 2  1010
20.
a.
b.
c.
d.
4.2  103 + 48.0  103 = 52.2  103 = 5.22  104
90  103 + 360  103 = 450  103 = 4.50  105
50  105  6  105 = 44  105 = 4.4  104
1.2  103 + 0.05  103  0.4  103 = 0.85  103 = 850
21.
a.
b.
c.
d.
e.
f.
(102)(103) = 105 = 100  103
(102)(103) = 101 = 10
(103)(106) = 1  109
(102)(105) = 1  103
(106)(10  106) = 10
(104)(108)(1028) = 1  1024
19.
2
b.
Chapter 1
22.
a.
b.
c.
d.
(50  103)(2  103) = 100  100 = 100
(2.2  103)(2  103) = 4.4  100 = 4.4
(82  106)(1.2  106) = 98.4
(30  104)(4  103)(7  108) = 840  101 = 8.40  103
23.
a.
b.
c.
d.
e.
f.
102/104 = 102 = 10  103
102/103 = 105 = 10  106
104/103 = 107 = 10  106
107/102 = 1.0  109
1038/104 = 1.0  1042
100 / 10 2 = 101/102 = 1  103
24.
a.
b.
c.
d.
(2  103)/(8  105) = 0.25  108 = 2.50  107
(4  103)/(4  106) = 4/4  109 = 1  109
(22  105)/(5  105) = 22/5  100 = 4.4
(78  1018)/(4  106) = 1.95  1025
25.
a.
c.
(102)3 = 1.0  106
(104)8 = 100.0  1030
26.
a.
b.
c.
d.
27.
a.
(2  102)2 = 4  104
(5  103)3 = 125  109
(4  103)(3  103)2 = (4  103)(9  104) = 36  101 = 360
((2  103)(0.8  104)(0.003  105))3 = (4.8  103)3 = (4.8)3  (103)3
= 110.6  109 = 1.11  1011
(103)2 = 1.0  106
(10 2 )(10 4 )
= 102/103 = 1.0  105
3
10
3 2
(10 ) (10 2 ) (10 6 )(10 2 ) 10 4

 4 = 1.0  108
4
4
10
10
10
3
4
(10 )(10 )
= 107/104 = 1.0  1011
4
10
b.
c.
d.
e.
(10 )(10 )  1  1 = 1.0  10
(10 ) 10  (10 )(10 ) 10
2 2
28.
a.
b.
c.
Chapter 1
(104)1/2 = 10.0  103
(107)9 = 1.0  1063
(1  104)3(102)/106 = (1012)(102)/106 = 1010/106 = 1.0  1016
2
f.
b.
d.
2
3
3
4
3
1
(3  10 2 ) 2 (10 2 )
= (9  104)(102)/(3  104) = (9  106)/(3  104) = 3  102 = 300
4
3  10
(4  10 4 ) 2 16  10 8
= 2  105

3
3
(20)
8  10
(6  104 ) 2
36  108
= 9.0  1012

2 2
4
(2  10 )
4  10
3
d.
(27  10 6 )1 / 3 3  10 2
= 1.5  107 = 150.0  109

5
5
2  10
2  10
e.
(4  10 3 ) 2 (3  10 2 ) (16  10 6 )(3  10 2 ) 48  10 8
= 24.0  1012


2  10  4
2  10  4
2  10  4
f.
(16  106)1/2(105)5(2  102) = (4  103)(1025)(2  102) = 8  1020 = 800.0  1018
g.
 (3  103 )3  1.60  102   (2  102 )(8  104 ) 


 

5 2
(7  10 )
2
1/ 2
(27  109 )(2.56  104 )(16  102 )1/ 2
49  1010
(69.12  105 )(4  101 ) 276.48  106


49  1010
49  1010
4
3
= 5.64  10 = 56.4  10

29.
Scientific:
Engineering:
30.
Scientific
Engineering:
4
a.
b.
2.05 × 101
5.04 × 104
c.
6.74 × 104
d.
4.60 × 102
a.
b.
20.46 × 100
50.42 × 103
c.
674.00 × 106
d.
46.00 × 103
a.
b.
5.0 × 102
4.5 × 101
c.
d.
1/32 = 0.03125 = 3.125 × 102
3.14159 = 3.142 × 100
a.
b.
50.0 × 103
0.045 × 103
c.
d.
31.25 × 103
3.142 × 100
Chapter 1
+2
31.
a.
6  104 = 0.06  10+6 = 0.06 × 10+6
3
3
b.
0.4  103 = 400  106 = 400 × 106
+3
2
c.
+3
50  105 = 5000  103 = 5  106 = 0.005  109 = 0.005 × 109
3
+2
3
3
+4
d.
+3
3
12  107 = 0.0012  103 = 1.2  106 = 1200  109 = 1200 × 109
4
+3
+3
3
32.
a.
0.05  100 s = 50  103 s = 50 ms
+3
+3
b.
2000  106 s = 2  103 s = 2 ms
3
3
c.
0.04  103 s = 40  106 s = 40 s
+3
+6
d.
8400  1012 s  0.0084  106 s = 0.0084 s
6
Chapter 1
5
+3
increase by 3
100
e.
100  103  103 m = 0.1  103 m = 0.1 km
3
33.
34.
35.
6
a.
 60 s 
1.5 min 
 = 90 s
1 min 
b.
 60 min   60 s 
2 × 102 h 
 = 72 s

 1 h  1 min 
c.
 1 s 
0.05 s   6  = 0.05  106 s = 50  103 s
10 s 
d.
 1 mm 
0.16 m  3  = 0.16  103 mm = 160 mm
10 m 
e.
 1 ns 
1.2  107 s  9  = 1.2  102 ns = 120 ns
10 s 
f.
1 min   1 h  1 day 
4  108 s 

 = 4629.63 days

 60 s   60 min   24 h 
a.
100 cm 
= 8000  103 cm = 8 cm
80  103 m 

 1m 
b.
 1 m   1 km 
= 60  105 km
60 cm 



100 cm  1000 m 
c.
 1 m 
12 × 103 m  6  = 12  103 × 10+6 m = 12 × 103 m
10 m 
d.
 1m  1m 
4
2
60 cm2 

 = 60  10 m
100 cm   100 cm 
a.
 1m 
100 in. 
 = 2.54 m
 39.37 in. 
b.
12 in.   1 m 
4 ft 

 = 1.22 m
 1 ft   39.37 in. 
Chapter 1
c.
d.
36.
37.
38.
39.
40.
41.
 4.45 N 
6 lb 
 = 26.7 N
 1 lb 
 1 N   1 lb 
60  103 dynes  5

 = 0.13 lb
10 dynes   4.45 N 
e.
 1 in.   1 ft 
150,000 cm 

 = 4921.26 ft
 2.54 cm  12 in. 
f.
 5280 ft  12 in.   1 m 
0.002 mi 


 = 3.22 m
 1 mi   1 ft   39.37 in. 
1 yd 
5280 ft 
 = 1760 yds
 3 ft 
 12 in.   1 m 
5280 ft 

 = 1609.35 m, 1.61 km
 1 ft   39.37 in. 
5280 ft,
60 mi 5280 ft
h
1 mi
12 in.
1 ft
1m
39.37 in.
1h
60 min
1000 m   39.37 in.   1 ft   1 mi 
10 km 




 1 km   1 m  12 in.   5280 ft 
d 6.214 mi
1 mi
=
,t= 
= 40.39 min
1 mi

6.5 min
6.5 min
1 min
60 s
= 26.82 m/s
= 6.214 mi
 3 ft  12 in. 
100 yds 

 = 3600 in  3600 quarters
1 yd   1 ft 
500 mi
= 8.33 h = 8 h:19.8 min
 60 mi/h
d 500 mi
= 6.67 h = 6h:40.2 min
75 mph: t = 
 75 mi/h
difference = 1h:28.6 minutes
60 mph:
t=
d

cm 

0.016 h   60 min   60 s   1 m  = 345.6 m
d = t = 600

s 

 1 h  1 min  100 cm 
Chapter 1
7
42.


 14 ft  1 step 
d = 86 stories 
 = 1605 steps

 story   9 ft 
 12 
d
d 1605 steps
 1 minute 
t  
 802.5 seconds
 = 13.38 minutes
2
steps
t

 60 seconds 
second
=
43.
44.
 14 ft 
 1 mile 
d = (86 stories) 
  1204 ft 
 = 0.228 miles
 5,280 ft 
 story 
min 10.22 min

= 44.82 min/mile
mile 0.228 miles
5 min
1 mile  5,280 ft  1056 ft

,

mile
5 min  1mile  minute
d
d
1204 ft
t 
= 1.14 minutes
 1056 ft
t
min
=
45.
a.
 1 Btu 
= 4.74  103 Btu
5 J

1054.35 J 
b.

1 m3
 1 gallon  
4 3
24 ounces 

 = 7.1  10 m

128 ounces   264.172 gallons 
c.
 86,400 s 
5
1.4 days 
 = 1.21  10 s
1
day


d.
 264.172 gallons   8 pints 
1 m3 
 1 gallon  = 2113.38 pints
1 m3



46.
6(4 × 2 + 8) = 96
47.
(42 + 6/5)/3 = 14.4
48.
2
5 
3
2
2
= 5.044
49.
MODE = DEGREES: cos 21.87 = 0.928
50.
MODE = DEGREES: tan1(3/4) = 36.87
51.
8
 14 ft 
distance = 86 stories 
 = 1204 ft
 story 
400 /(6  10 / 5) = 7.071
2
Chapter 1
52.
205  106
53.
1.20  1012
54.
6.667  106 + 0.5  106 = 7.17  106
Chapter 1
9
Chapter 2
1.

2.
a.
F= k
Q1Q2 (9  109 )(1 C)(2 C)
= 18  109 N

r2
(1 m) 2
b.
F=k
Q1Q2
(9  109 )(1 C)2 C 
=
= 2  109 N
r2
(3 m) 2
c.
F= k
Q1Q2 (9  109 )(1 C)(2 C)
= 0.18  109 N

2
2
r
(10 m)
d.
Exponentially,
a.
r = 1 mi:
3.
r3 10 m
F
18  109 N
= 10 while 1 
= 100

r1 1 m
F2 0.18  109 N
 5280 ft  12 in.   1 m 
1 mi 


 = 1609.35 m
 1 mi   1 ft   39.37 in. 
kQ1Q2 (9  109 )(8  106 C)(40  10-6 C) 2880  103
F=


r2
(1609.35 m)2
2.59  106
= 1.11 N
b.
r =10 ft:
12 in.   1 m 
10 ft 

 = 3.05 m
 1 ft   39.37 in. 
kQ1Q2 2880  103 2880  103
F=
= 0.31 N


9.30
r2
(3.05 m) 2
c.
1 in.  1 m 
= 1.59 mm
16  39.37 in. 
kQ1Q2
2880  103
2880  103


 1138.34  103 N
r2
(1.59  103 m) 2 2.53  106
= 1138.34 kN
F=
4.

5.
Q1 = Q2 = Q; F1 
6.
F=
10
kQ1Q2
r
r2
F1r12
r12
kQ 2
kQ 2 k  F1r12 
2
;
and
Q
F




F

F1


2
2
k
r12
r22
r22  k 
r22
kQ1Q2

F
(9  109 )(20  106 ) 2
= 10 mm
3.6  104
Chapter 2
7.
F=
a.
b.
kQ1Q2
kQ Q
 1.8  1 22  kQ1Q2 = 4(1.8) = 7.2
2
r
(2 m)
kQ1Q2
7.2
= 72 mN

2
r
(10) 2
Q1/Q2 = 1/2  Q2 = 2Q1
7.2 = kQ1Q2 = (9  109)(Q1)(2Q1) = 9  109 2Q12
F=
 
7 .2
7 .2
 Q12  Q1 
= 20 C
9
18  10
18  109
Q2 = 2Q1 = 2(2  105 C) = 40 C
W
1.2 J

= 60 kV
Q 20 C
8.
V=
9.
W = VQ = (60 V)(8 mC) = 0.48 J
10.
Q=
 6.242  1018 electrons 
W 120  J
15

= 6 mC 
 = 37.45 × 10 electrons


V
20 mV
1C
11.
Q=
W 72 J
=8C

V
9V
12.
a.
b.
W = QV = (1 × 1012 electrons)(40 V) = 40 × 1012 eV
1C


40 × 1012 eV 
18
 = 6.41 μJ
6.242
10
electrons


13.
I=
Q 12 mC

= 4.29 mA
t
2.8 s
14.
I =
Q
312 C
= 2.60 A

t (2)(60 s)
15.
Q = It = (40 mA)(0.8)(60 s) = 1.92 C
16.
Q = It = (250 mA)(1.2)(60 s) = 18.0 C
17.
t=
18.
Q 6 mC

=3s
I
2 mA
1C


21.847  1018 electrons 
18
 = 3.5 C
 6.242  10 electrons 
Q 3.5 C
I=

= 0.29 A
t
12 s
Chapter 2
11
19.
Q = It = (4 mA)(90 s) = 360 mC
 6.242  1018 electrons 
18
360 mC 
 = 2.25  10 electrons
1
C


20.
I=
21.
22.
Q
86 C
= 1.194 A > 1 A (yes)

t
(1.2)(60 s)
1C


0.84  1016 electrons 
18
 = 1.346 mC
6.242

10
electrons


Q 1.346 mC
I=

= 22.43 mA
t
60 ms
a.
Q = It = (2 mA)(0.01 s) = 2  1011 C
 6.242  1018 electrons   1 ¢ 
2  1011 C 


1C

  electron 
= 1.25  108 ¢ = $1.25  106 = 1.25 million
Q = It = (100 A)(1.5 ns) = 1.5  1013 C
 6.242  1018 electrons   $1 
1.5  1013 C 

 = 0.94 million
1C

  electron 
(a) > (b)
b.
23.
24.
Q = It = (200  103 A)(30 s) = 6 C
W 40 J
= 6.67 V
V=

Q 6C
 420 C 
Q = It = 
 (0.5 min) = 210 C
 min 
W
742 J
= 3.53 V
V=

Q 210 C
W 0.4 J

= 0.0167 C
V
24 V
Q 0.0167 C
I=

= 3.34 A
t 5  103 s
25.
Q=
26.
I=
27.
Ah = (0.8 A)(75 h) = 60.0 Ah
28.
t(hours) =
12
Ah rating 200 Ah
=5A

t (hours)
40 h
Ah rating 32 Ah

= 25 h
I
1.28 A
Chapter 2
29.
 60 min   60 s 
6
40 Ah(for 1 h): W1 = VQ = VIt = (12 V)(40 A)(1 h) 
 1 min  = 1.728  10 J
1
h



60
min
60
s



6
60 Ah(for 1 h): W2 = (12 V)(60 A)(1 h) 
 1 min  = 2.592  10 J
1
h



Ratio W2/W1 = 1.5 or 50% more energy available with 60 Ah rating.
1 min   1 h 
= I(16.67  103 h)
For 60 s discharge: 40 Ah = It = I 60 s 



 60 s   60 min 
40 Ah
and I =
= 2400 A
16.67  10-3 h
1 min   1 h 
= I(16.67  103 h)
60 Ah = It = I 60 s 



 60 s   60 min 
60 Ah
= 3600 A
and I =
16.67  10-3 h
I2/I1 = 1.5 or 50 % more starting current available at 60 Ah
30.
0.75(18 Ah) = 13.5 Ah   250 mA
31.
(18 Ah  15.5 Ah)/18 Ah × 100% = 13.89%
32.
At 100 mA, discharge time  120 H; At 25 mA, discharge time  425 h;
 300 h more at 25 mA
33.
I=
3 Ah
= 500 mA
6.0 h
 60 min   60 s 
Q = It = (500 mA)(6 h) 

 = 10.80 kC
 1 h  1 min 
W = QV = (10.8 kC)(12 V)  129.6 kJ
34.

35.

36.

37.
a.
b.
c.
Chapter 2
 2.54 cm 
= 1.27 cm
0.5 in 
 1 in 
 30 kV 
= 38.1 kV
1.27 cm 
 cm 
 270 kV 
1.27 cm 
= 342.9 kV
 cm 
342.9 kV:38.1 kV = 9:1
13
38.

39.

40.

41.

14
Chapter 2
Chapter 3
1.
a.
0.5 in. = 500 mils
b.
1000 mils 
0.02 in. 
 = 20 mils
 1 in. 
c.
1
1000 mils 
in. = 0.25 in. 
 = 250 mils
4
 1 in. 
d.
e.
2.
3.
 39.37 in  1000 mils 
10 mm = 10 × 103 m 
= 393.7 mils
 1 m   1 in 
3
12 in.  10 mils 
0.01 ft 
 = 120 mils

 1 ft   1 in. 
f.
 1 in.  1000 mils 
0.1 cm 

 = 39.37 mils
 2.54 cm   1 in. 
a.
ACM = (30 mils)2 = 900 CM
b.
0.016 in. = 16 mils, ACM = (16 mils)2 = 256 CM
c.
1"
= 0.125" = 125 mils, ACM = (125 mils)2 = 15.63  103 CM
8
d.
 1 in.  1000 mils 
2
3
1 cm 
  1 in.  = 393.7 mils, ACM = (393.7 mils) = 155  10 CM
2.54
cm



e.
12 in.  1000 mils 
2
3
0.02 ft 
  1 in.  = 240 mils, ACM = (240 mils) = 57.60  10 CM
1
ft



f.
 39.37 in.  1000 mils 
= 157.48 mils, ACM = (157.48 mils)2 = 24.8 × 103 CM
4 × 103 m 



 1 m   1 in 
ACM = (dmils )2  dmils =
ACM
a.
d = 1600 CM = 40 mils = 0.04 in.
b.
d=
820 CM = 28.64 mils = 0.029 in.
c.
d=
40,000 CM = 200 mils = 0.2 in.
Chapter 3
15
d.
d=
625 CM = 25 mils = 0.025 in.
e.
d=
6.25 CM = 2.5 mils = 0.0025 in.
f.
d=
3  103 CM = 54.77 mils = 0.055 in.
4.
0.02 in. = 20 mils, ACM = (20 mils)2 = 400 CM
(200 )
l
= 5.19 
R =  = (10.37)
400 CM
A
5.
a.
A= 
 80 
l
 = 544 CM
 17
R
 2 .5  
b.
d=
ACM  544 CM = 23.32 mils = 23.3  103 in.
6.
7.
1 "
= 0.03125" = 31.25 mils, ACM = (31.25 mils)2 = 976.56 CM
32
l
RA (2.2 )(976.56 CM)
= 3.58 ft
R= l=

R

600
a.
ACM = 
d=
l
(10.37)(300)
=
= 942.73 CM
3.3 
A
942.73 CM = 30.70 mils = 30.7  103 in.
b.
larger
c.
smaller
8.
 =
RA (500 )(94 CM)
= 47  nickel

l
1000
9.
a.
b.
c.
16
1/32 = 0.03125 = 31.25 mils, ACM = (31.25 mils)2 = 976.56 CM
l
RA (3.12 )(976.56 CM)
= 293.82 ft
l 

R=
A

10.37
293.82 1000
(5)(293.82)
= 1.47 lbs

x
x
5 lb
1000
9
9
C  32  (40)  32 = 40°
5
5
9
9
105° C: F = C  32  (105)  32 = 221°
5
5
F° = 40°  221°
40° C: F =
Chapter 3
10.
a.
3 
 0.375 = 375 mils
8
4.8 = 4800 mils
 4 /  CM 
6
A = (375 mils)(4800 mils) = 1.8 × 106 sq. mils 
 = 2.29 × 10 CM
1
sq
mil


b.
11.
a.
1 
= 0.083 in. = 83 mils
12
ACM = = (83 mils)2 = 6.89 × 103 CM
2.29  106 CM
(#12)
= 332.37 wires
6.89  103 CM
3" = 3000 mils, 1/2" = 0.5 in. = 500 mils
Area = (3  103 mils)(5  102 mils) = 15  105 sq. mils
 4 /  CM 
5
15  105 sq mils 
 = 19.108  10 CM
 1 sq mil 
R= 
b.
l
(10.37)(4)
=
= 21.71 
A 19.108  105 CM
l
(17)(4)
=
= 35.59 
A 19.108  105 CM
Aluminum bus-bar has almost 64% higher resistance.
R= 
12.
l2 = 2l1, A2 = A1/4, 2 = 1
 2l2
R2
l A
2l A
A
 2  22 1  1 1 =8
1l1
R1
1l1 A2 l1 A1 / 4
A1
and R2 = 8R1 = 8(0.2 ) = 1.6 
R = 1.6   0.2  = 1.4 
13.
A=
d 2
4
d=
4A


4(0.04 in.2 )

= 0.2257 in.
dmils = 225.7 mils
ACM = (225.7 mils)2 = 50,940.49 CM
l
1 1
R1
lA
lA
A1

 11 2  1 2
(1 = 2)
R2  l2
 2l2 A1 l2 A1
2
A2
R l A (800 m)(300 ft)(40,000 CM)
= 942.28 m
and R2 = 1 2 1 
(200 ft)(50,940.49 CM)
l1 A2
Chapter 3
17
14.
a.
b.
15.
#12 = 6,529.9 CM, #14 = 4,106.8 CM
6,529.9 CM  4,106.8 CM
 100% = 59% larger
4,106.8 CM
#12 20 A
#12 6,529.9 CM

= 1.33,

= 1.59
#14 4,106.8 CM
#14 15 A
Imax ratio = 1.33 vs Area ratio = 1.59
1.59  1.33
 100% = 19.55% higher ratio for area
1.33
a.
#9 13,094 CM

= 2 yes
#12 6,529.9 CM
b.
#0 105,530 CM

= 16.16 yes
#12 6,529.9 CM
#0 150 A

= 7.5
#12 20 A
16.
a.
b.
17.
18.
18
#10 10,381 CM

 10.16  10 yes
# 20 1,021.5 CM
# 20 1,021.5 CM

= 103.28
# 40
9.89 CM
yes  100
l (10.37)(30) 311.1 CM
= 51,850 CM  #3


R
6 m
6  103
but 110 A  #2
a.
A= 
b.
A= 
a.
A/CM = 230 A/211,600 CM = 1.09 mA/CM
b.


1.09 mA  1 CM  1000 mils  1000 mils 
= 1.39 kA/in.2





CM  sq mils   1 in.   1 in. 
4

c.
 1 in.2 
2
5 kA 
 = 3.6 in.
1.39 kA 
l (10.37)(30) 311.1 CM
= 103,700 CM  #0


3 m
R
3  10 3
Chapter 3
19.
20.
234.5  10 234.5  80
,

2
R2
R2 =
(314.5)(2 )
= 2.57 
244.5
236  0 236  100

R2
0.02 
(0.02 )(336)
= 0.028 
R2 =
236
5
5
(F  32)  (32  32) = 0 (=32F)
9
9
5
C = (68  32)  20 (=68F)
9
234.5  20 234.5  0

4
R2
(234.5)(4 )
= 3.69 
R2 =
254.5
21.
C=
22.
a.
b.
c.
d.
Chapter 3
5
5
F  32  (70  32) = 21.11°
9
9
5
°C = (60  32) = 15.56°
9
234.5  21.11 234.5  15.56

0.025 
R2
(250.06)(0.025 )
= 24.46 mΩ
R2 =
255.61
°C =
5
(50  32) = 10°
9
234.5  21.11 234.5  10

0.025 
R2
(244.5)(0.025 )
= 23.91 mΩ
R2 =
255.61
°C =
Part a: 25 mΩ  24.46 mΩ = 0.54 mΩ
Part b: 24.45 mΩ  23.91 mΩ = 0.55 mΩ
Linear 40°F  23.91 mΩ  0.55 mΩ = 23.36 mΩ
5
(30  32) = 34.44°
9
234.5  21.11 234.5  34.44

25 m
R2
(25 m)(200.06)
= 19.57 mΩ
R2 =
255.61
Yes, 25 mΩ  19.57 mΩ = 5.43 mΩ
°C =
19
e.
23.
a.
b.
24.
5
(120  32) = 48.89°
9
234.5  21.11 234.5  48.89

25 m
R2
(25 m)(283.39)
= 27.72 mΩ
R2 =
255.61
Yes, 2.72 mΩ
°C =
234.5  4 234.5  t2
,

1
1 .1 
234.5  4 234.5  t2
,

1
0.1 
t2 = 27.85C
t2 = 210.65C
68F  20C
a.
234.5+20 234.5  T2

1
2
2(254.5)
 234.5  T2
1
T2  274.5C
b.
#10 = 0.9989 /1000
c.
d mils  ACM  10,381 CM  101.89 mils
din = 0.102 in 
25.
26.
20
1 
10
1
1
1
= 0.003929  0.00393


Ti  20C 234.5  20 254.5
a.
20 =
b.
R = R20[1 + 20(t  20C)]
1  = 0.8 [1 + 0.00393(t  20)]
1.25 = 1 + 0.00393t  0.0786
1.25  0.9214 = 0.00393t
0.3286 = 0.00393t
0.3286
t=
= 83.61C
0.00393
R = R20[1 + 20(t  20C)]
= 0.4 [1 + 0.00393(16  20)] = 0.4 [1  0.01572] = 0.39 
Chapter 3
27.
Table: 1000 of #12 copper wire = 1.588  @ 20C
5
5
C = (F  32) = (115  32) = 46.11C
9
9
R = R20[1 + 20(t  20C)]
= 1.588 [1 + 0.00393(46.11  20)]
= 1.75 
28.
R =
29.
R =
30.
a.
31.
10 k  3.5 k = 6.5 k
32.
6.25 k and 18.75 k
33.

34.
a.
b.
c.
d.
e.
820   5%, 820   41 , 779   861 
220   10%, 220   22 , 198   242 
91 k  20%, 91 k  18.2 k, 77.8 k  109.2 k
9.1 k  5%, 9100   455 , 8,645   9,555 k
3.9 MΩ  20%, 3.9 MΩ  0.78 MΩ, 3.12 MΩ  4.68 M
35.
a.
b.
c.
d.
68  = Blue, Gray, Black, Silver
0.33  = Orange, Orange, Silver, Silver
22 k = Red, Red, Orange, Silver
5.6 M = Green, Blue, Green, Silver
36.
a.
10   20%
8
12
15 Ω ± 20%
12
18
10 Ω ± 10%
9
11
15 Ω ± 10%
13.5
Rnominal
22 
(PPM)( T) =
(200)(65  20) = 0.198 
6
10
106
R = Rnominal + R = 22.198 
Rnominal
100 
(PPM)( T) =
(100)(50  20) = 0.30 
6
10
106
R = Rnominal + R = 100  + 0.30  = 100.30 
b.
37.
38.
2 times larger
b.
4 times larger
no overlap, continuance
no overlap
16.5
470 Ω ± 10% = 470 Ω ± 47 Ω
= 423 Ω  517 Ω
Yes
No change
Chapter 3
21
39.
a.
b.
c.
d.
621 = 62  101  = 620  = 0.62 k
333 = 33  103  = 33 k
Q2 = 3.9  102  = 390 
C6 = 1.2  106  = 1.2 M
40.
a.
G=
1
1
= 8.33 mS

R 120 
b.
G=
1
= 0.25 mS
4 k
c.
41.
42.
43.
1
= 0.46 S
2.2 M
Ga > Gb > Gc vs. Rc > Rb > Ra
G=
a.
Table 3.2, /1000 = 1.588 
1
1
= 629.72 mS
G= 
R 1.588 
A 6529.9 CM (Table 3.2)
= 629.69 mS (Cu)
or G =

l
(10.37)(1000)
b.
G=
6529.9 CM
= 384.11 mS (Al)
(17)(1000)
a.
G1 =
1
1
1
= 100 mS, G2 =
= 50 mS, G3 =
= 10 mS
10 
20 
100 
b.
G2:G1 = 50 mS: 100 mS = 1:2 whereas R2:R1 = 20 Ω:10 Ω = 2:1. The rate of change is
the same although one is increasing and the other decreasing.
c.
inverse  linear
l
2
5
 2
A1 = A1, l2 = 1   l1  1 , 2 = 1
3
3
3
 3
A
 l1 
1 1
 3  A1 1
G1
l1
l A

 22 1  

A2 1l1 A2
5  5
G2

2
l1  A1 
l2
3 
A2 = 1
G2 = 5G1 = 5(100 S) = 500 S
44.

45.

46.

22
Chapter 3
47.

48.
1
 2.54 cm 
in. = 0.083 in. 
 = 0.21 cm
12
 1 in. 
d2
(3.14)(0.21 cm) 2
= 0.035 cm2
4
4
RA (2 )(0.035 cm 2 )

l=
= 40,603 cm = 406.03 m

1.724  106
A=
49.
a.

1 "  2.54 cm 
 2.54 cm 
= 1.27 cm, 3 in. 


 = 7.62 cm
2  1" 
 1 in . 
12 in.   2.54 cm 
4 ft 

 = 121.92 cm
 1 ft   1 in. 
l (1.724 x 10 -6)(121.92 cm)
R= 
= 21.71 Ω
A
(1.27 cm)(7.62 cm)
 (2.825  106 )(121.92 cm)
= 35.59 Ω

A
(1.27 cm)(7.62 cm)
b.
R= 
c.
increases
d.
decreases

= 100  d =

Rs =
51.
R l (150 )(1/ 2 in.)
R = Rs l  w = s =
= 0.15 in.
R
500 
w
52.
a.
d = 1 in. = 1000 mils
ACM = (103 mils)2 = 106 CM
RA (1 m)(106 CM)
1 =
= 1 CM-/ft

l
103 ft
b.
1 in. = 2.54 cm
 d 2  (2.54 cm)2
A=

= 5.067 cm2
4
4
12 in.   2.54 cm 
l = 1000 ft 

 = 30,480 cm
 1 ft   1 in. 
RA (1 m)(5.067 cm 2 )

2 =
= 1.66  107 -cm
l
30,480 cm
Chapter 3
d
100
=
250  106
= 2.5 cm
100
50.
23
c.
k=
 2 1.66  107 -cm
= 1.66  107

1 CM- / ft
1
53.

54.

55.

56.

57.

58.
a.
50C specific resistance  105 -cm
50C specific resistance  500 -cm
200C specific resistance  7 -cm
b.
negative
c.
No
d.
=
a.
Log scale:
10 fc  3 k
100 fc  0.4 k
b.
negative
c.
no—log scales imply linearity
d.
1 k   30 fc
10 k   2 fc
R
10 k   1 k 
=
= 321.43 /fc
30 fc  2 fc
fc
59.
  cm 300  30 270   cm
 3.6 -cm/C


125  50
75 C
T
and
60.
24
a.
@ 0.5 mA, V  195 V
@ 1 mA, V  200 V
@ 5 mA, V  215 V
b.
Vtotal = 215 V  195 V = 20 V
c.
5 mA:0.5 mA = 10:1
compared to
215 V: 200 V = 1.08:1
R
= 321.43 /fc
 fc
Chapter 3
Chapter 4
1.
V = IR = (5.6 mA)(220 ) = 1.23 V
2.
I=
V 24 V

= 3.53 A
R 6.8 
3.
R=
V
24 V
= 16 k
=
I 1.5 mA
4.
I=
V
12 V

= 300 A
R 40 103 
5.
V = IR = (3.6 A)(0.02 M) = 0.072 V = 72 mV
6.
I=
V 120 V

= 2.4 mA
R 50 k
7.
R=
V 120 V
= 54.55 
=
I 2.2 A
8.
I=
V 120 V

= 15 mA
R 8 k
9.
R=
V 120 V
=
= 28.57 
I 4.2 A
10.
R=
V
4.5 V
= 36 
=
I 125 m A
11.
R=
V 24 mV
= 1.2 k
=
I 20  A
12.
V = IR = (12 A)(0.5 ) = 6 V
13.
a.
R=
V 120 V
= 12.63 
=
I 9.5 A
b.
 60 min   60 s 
t = 2 h

 = 7200 s
 1 h   1 min 
W = Pt = VIt
= (120 V)(9.5 A)(7200 s)
= 8.21  106 J
14.
V = IR = (2.4 A)(3.3 M) = 7.92 V
15.

Chapter 4
25
16.
b.
17.

18.

19.

20.
P=
W

t
21.
t=
W 640 J

= 16 s
P 40 J/s
22.
a.
b.
(0.13 mA)(500 h) = 65 mAh
540 J
540 J
= 2.25 W

60 s
240 s
4 min
1 min
 60 min   60 s 
8 h

 = 28,800 s
 1 h   1 min 
W = Pt = (2 W)(28,000 s) = 57.6 kJ
kWh =
(2 W)(8 h)
= 16  103 kWh
1000
23.
P = VI = (3 V)(1.4 A) = 4.20 W
W
12 J

= 2.86 s
t=
P 4.2 W
24.
P = EI = (6 V)(750 mA) = 4.5 W
25.
P = I2R = (7.2 mA)2 4 k = 207.36 mW
26.
P = I2R  I =
27.
I=
28.
29.
26
P

R
240 mW
= 10.44 mA
2.2 k
P
2W
= 129.10 mA

R
120 
V = IR = (129.10 mA)(120 ) = 15.49 V
E
22 V

= 1.31 mA
R 16.8 k
P = I2R = (1.31 mA)2 16.8 k = 28.83 mW
  60 min   60 s  
W = Pt = (28.83 mW) 1 h 

  = 103.79 J
  1 h  1 min  
I=
E=
P 324 W

= 120 V
I
2.7 A
Chapter 4
30.
I=
P

R
1W
= 461.27 A
4.7 M
no
PR  (42 mW)(2.2 k)  92.40 = 9.61 V
31.
V=
32.
P = VI, I =
33.
34.
P 100 W

= 0.833 A
V 120 V
V
120 V
R= 
= 144.06 
I 0.833 A
P 450 W

= 120 V
I 3.75 A
V 120 V
R= 
= 32 
I 3.75 A
V=
P 0.4  103 W
= 0.13 mA

E
3V
a.
P = EI and I =
b.
Ah rating = (0.13 mA)(500 h) = 66.5 mAh
P
100 W

 5  103 = 70.71 mA
R
20 k
V = PR  (100 W)(20 k) = 1.42 kV
35.
I=
36.
P = EI = (220 V)(30 A) = 6.6 kW
6.6 kW
= 8.85 HP
PHP =
746 W/HP
37.
a.
 V 2   12 V  2
W = Pt = 
t  
 60 s = 86.4 J
 R   10  
b.
Energy doubles, power the same
39.
40.
 1

4 weeks 
12 h  3

 [5 months] = 260 h
week  1 month 


(230 W)(260 h)
kWh =
= 59.80 kWh
1000
a.
Chapter 4
 60 min   60 s 
= 2.16 × 106 Ws
W = Pt = (60 W)(10 h) 
 1 h   1 min 
27
41.
42.
43.
44.
28
b.
1 Ws = 1 J  2.16 × 106 J
c.
W = Pt = (60 W)(10 h) = 600 Wh
d.
600 Wh
= 0.6 kWh
1000 W/1 kWh
e.
Cost = (0.6 kWh)(11 ¢/kWh) = 6.6 ¢
a.
kWh =
b.
I=
c.
Plost = Pi  Po = Pi  Pi = Pi(1  ) = 120 kW(1  0.82) = 21.6 kW
Pt
(21.6 kW)(10 h)

= 216 kWh
kWhlost =
1000
1000
Pt
(1000)(kWh) (1000)(1200 kWh)
P

= 120 kW
1000
P
10 h
P 120  103 W

= 576.92 A
E
208 V
$1.00
= 9.09
11¢
Pt
(kWh)(1000) (9.09)(1000)
kWh =

= 36.36 h
t
1000
P
250 W
#kWh =
t=
(kWh)(1000) (11.11)(1000)

= 2.32 h
P
4800
a.
$74
= $2.39/day
31 days
b.
$2.39 / day
= 16¢/h
15 h/day
c.
16¢
= 1.45 kWh
11¢/kWh
d.
1.45 kWh
= 24.16  24 bulbs
60 W
e.
no
$1.00
= 9.09 kWh
11¢/kWh
9.09 kWh
= 48.61 h
187 W
Chapter 4
45.
t = 5 h/day(365 days) = 1825 h
P  t (339 W)(1825 h)
= 618.68 kWh
kWh =

1000
1000
Cost = (618.68 kWh)(11¢/kWh) = $68.05
P  t (213 W)(1825 h)
= 388.73 kWh

kWh =
1000
1000
Cost = (388.73 kWh)(11¢/kWh) = $42.76
Cost Savings = $68.05  $42.76 = $25.29
46.
P  t (78 W)(4 h/day)(31 days)
= 9.67 kWh

1000
1000
Cost = (11¢/kWh)(9.67 kWh) = $1.06
47.
a.
P = EI = (120 V)(100 A) = 12 kW
b.
 746 W 
PT = 5 hp 
 + 3000 W + 2400 W + 1000 W
 hp 
kWh =
= 10,130 < 12,000 W (Yes)
c.
W = Pt = (10.13 kW)(2 h) = 20.26 kWh
(1600 W)(6 h)  (1200 W)(1/4 h)  (4800 W)
48.
kWh =
1h
1
h  (900 W) 10 min
2
60 min)
 (200 W)(2 h)  (50 W)(3.5 h)
1000
9600 Wh +300 Wh  2400 Wh + 150 Wh  400 Wh + 175 Wh
=
= 13.025 kWh
1000
(13.025 kWh)(11¢/kWh) = 1.43¢
(200 W)(4 h)  (6)(60 W)(6 h) + (1200 W) 20 min
49.
kWh =
1h
60 min
1
6
 (175 W)(3.5 h) + (250 W) 2 h  (30 W)(8 h)
1000
800 Wh  2160 Wh + 400 Wh + 612.5 Wh + 541.67 Wh + 240 Wh
=
= 4.754 kWh
1000
(4.754 kWh)(11¢/kWh) = 52.29¢
50.
=
51.
=
Po
 100% 
Pi
746 W
hp
373
 100% 
 100% = 90.98%
410 W
410
(0.5 hp)
P (1.8 hp)(746 W/hp)
Po
, Pi = o 
= 1960.29 W
Pi

0.685
P 1960.29 W
Pi = EI, I = i 
= 16.34 A
E
120 V
Chapter 4
29
52.
=
Po
(0.8 hp)(746 W/hp)
596.8
 100% 
 100% 
 100% = 67.82%
(4 A)(220 V)
880
Pi
53.
a.
Pi = EI = (120 V)(1.8 A) = 216 W
Pi = Po + Plost, Plost = Pi  Po = 216 W  50 W = 166 W
b.
% =
Po
Po
50 W
 100% =
 100% = 23.15%
216 W
Pi
I=
Po (3.6 hp)(746 W/hp)
= 16.06 A
=
E
(0.76)(220 V)
54.
Pi = EI =
55.
a.
Pi = P o =
b.
Pi = EI = 1657.78 W
(110 V)I = 1657.78 W
1657.78 W
= 15.07 A
I=
110 V
c.


Po
(2 hp)(746 W/hp)
= 1657.78 W
0.9
(2 hp)(746 W/hp)
= 2131.43 W

0.7
Pi = EI = 2131.43 W
(110 V)I = 2131.43 W
2131.43 W
= 19.38 A
I=
110 V
Pi =
Po
=
(15 hp)(746 W/hp)
= 12,433.33 W

0.9
12, 433.33 W
I = Pi =
= 56.52 A
E
220 V
56.
Pi =
57.
T = 1  2
=
0.75 = 0.85  2
2 = 0.88
58.
T = 1  2 = (0.87)(0.75) = 0.6525  65.25%
59.
T = 1  2 = 0.78 = 0.92
2 =
30
0.78
= 0.867  86.7%
0.9
Chapter 4
60.
a.
T = 1  2  3 = (0.93)(0.87)(0.21) = 0.170  17%
b.
T = 1  2  3 = (0.93)(0.87)(0.80) = 0.647  64.7%
64.7%  17%
 100% = 280.59%
17%
61.
T =
Po
= 1  2 = 1  21 = 212
Pi
12 
Po
 1 =
2 Pi
Po
128 W
= 0.4

2 Pi
2(400 W)
2 = 21 = 2(0.4) = 0.8
 2 = 40%, 2 = 80%
Chapter 4
31
Chapter 5
1.
a.
b.
c.
d.
e.
f.
E and R1
R1 and R2
E1, E2, and R1
E1 and R1; E2, R3 and R4
R3, R4 and R5; E and R1
R2 and R3
2.
a.
b.
c.
d.
RT = 0.1 k + 0.39 k + 1.2 k + 6.8 kΩ = 8.49 k
RT = 1.2  + 2.7  + 8.2  = 12.1 
RT = 8.2 k + 10 k + 9.1 k + 1.8 k + 2.7 k = 31.8 k
RT = 47  + 820  + 91  + 1.2 k = 2158.0 
3.
a.
b.
RT = 1.2 k + 1 k + 2.2 k + 3.3 k = 7.7 k
RT = 1 k + 2 k + 3 k + 4.7 k + 6.8 k = 17.5 k
4.
a.
b.
c.
1 M
100 , 1 k
RT = 100  + 1 k + 1 M + 200 k = 1.2011 M vs. 1.2 M for part b.
5.
a.
b.
RT = 10  + 33  + 56 Ω, Reading = 99 
RT = 2.2 k + 0.82 kΩ + 1.2 k + 3.3 k, Reading = 7.52 k
6.
a.
b.
RT = 129 k = R + 56 k + 22 k + 33 k, Reading = 18 k
RT = 103 k = 24 k + R1 + 43 k + 2R1 = 67 k + 3R1, R1 = 12 k
R2 = 24 k
7.
a.
b.
c.
1.2 k
0

8.
a.
RT = 10  + 12  + 18  = 40 
E 72 V
= 1.8 A
Is =

RT 40 
V1 = I1R1 = (1.8 A)(10 ) = 18 V, V2 = I2R2 = (1.8 A)(12 ) = 21.6 V,
V3 = I3R3 = (1.8 A)(18 ) = 32.4 V
Ps = EIs = (72 V)(1.8 A) = 129.6 W
P18 = V3I3 = (32.4 V)(1.8 A) = 58.32 W
b.
c.
d.
e.
9.
a.
b.
c.
32
the most: R3, the least: R1
R3, RT = 1.2 k + 6.8 k + 82 k = 90 k
E
45 V
= 0.5 mA
Is =

RT 90 k
V1 = I1R1 = (0.5 mA)(1.2 k) = 0.6 V, V2 = I2R2 = (0.5 mA)(6.8 k) = 3.4 V,
V3 = I3R3 = (0.5 mA)(82 k) = 41 V, results agree with part (a)
Chapter 5
10.
d.
e.
Ps = EIs = (72 V)(1.8 A) = 129.6 W
P18Ω = V3I3 = (32.4 V)(1.8 A) = 58.32 W
a.
RT = 12 k + 4 k + 6 k = 22 k
E = IRT = (4 mA)(22 k) = 88 V
RT = 12  + 22  + 82  + 10  = 126 
E = IRT = (500 mA)(126 ) = 63 V
b.
11.
a.
a.
b.
c.
d.
b.
a.
b.
c.
d.
12.
13.
V 5.2 V

=4A
R 1.3 
E = IRT = (4 A)(9 ) = 36 V
RT = 9  = 4.7  + 1.3  + R,
V4.7  = (4 A)(4.7 ) = 18.8 V
V1.3  = (4 A)(1.3 ) = 5.2 V
V3  = (4 A)(3 ) = 12 V
I=
R=3
V
6.6 V

= 3 mA
R 2.2 k
V3.3 k = (3 mA)(3.3 k) = 9.9 V
I=
E = 6.6 V + 9 V + 9.9 V = 25.5 V
V
9V

= 3 k
R=
I 3 mA
V2.2 k = 6.6 V, V3 k = 9 V, V3.3 k = 9.9 V
1
1
E
36 V
= 8.18 mA, Vm = E  (36 V) = 18 V

RT 4.4 k
2
2
a.
Im =
b.
RT = 1 k + 2.4 k + 5.6 k = 9 k
E
22.5 V
= 2.5 mA, Vm = 2.5 mA(2.4 k + 5.6 k) = 20 V
Im =

RT
9 k
c.
V3.3kΩ =
a.
RT = 3 k + 1 k + 2 k = 6 k
E 120 V
= 20 mA
Is =

RT
6 k
VR1 = (20 mA)(3 k) = 60 V
3.3 k(12 V)
= 8.8 V
4.5 k
Vm = 12 V  8.8 V = 3.2 V
12 V
Im =
= 2.67 mA
4.5 k
VR2 = (20 mA)(1 k) = 20 V
VR3 = (20 mA)(2 k) = 40 V
Chapter 5
33
b.
PR1 = I12 R1 = (20 mA)2  3 k = 1.2 W
PR2 = I 22 R2 = (20 mA)2  1 k = 0.4 W
PR3 = I 32 R3 = (20 mA)2  2 k = 0.8 W
c.
PT = PR1  PR2  PR3 = 1.2 W + 0.4 W + 0.8 W = 2.4 W
d.
PT = EIs = (120 V)(20 mA) = 2.4 W
e.
the same
f.
R1  the largest
g.
dissipated
h.
R1: 2 W, R2 : 1/2 W, R3: 1 W
14.
P = 21 W = (1 A)2R, R = 21 
V1 = I1R1 = (1 A)(2 ) = 2 V, V2 = I2R2 = (1 A)(1 ) = 1 V
V3 = I3R3 = (1 A)(21 ) = 21 V
E = V1 + V2 + V3 = 2 V + 1 V + 21 V = 24 V
15
P = 8 W = I21 , I = 8 = 2.828 A
P = 16 W = I2R1 = (2 .828 A)2R1, R1 = 2 
RT = 32  = 2  + R2 + 1  = 3  + R2, R2 = 29 
E = IRT = (2.828 A)(32 ) = 90.5 V
16.
a.
 1 
RT = NR1 = 8 28   = 225 
 8 
120 V
E
I=
= 0.53 A

RT 225 
2
34
b.
 8   1   64  225 
P = I2R =  A   28    

 =8W
 15   8   225  8 
c.
 8  225 
V = IR =  A 
  = 15 V
 15  8

d.
All go out!
Chapter 5
17.
Ps = PR1  PR2  PR3
EI = I2R1 + I2R2 + 24
(R1 + R2)I2  EI + 24 = 0
6I2  24 I + 24 = 0
I2  4 I + 4 = 0
 (4)  (4) 2  4(1)(4) 4  16  16 4

 =2A
2(1)
2
2
24 
=6
P = 24 W = (2 A)2R, R =
4
I=
18.
a.
b.
c.
Vab + 4 V + 24 V  12 V = 0, Vab = 28 V + 12 V = 16 V
Vab + 4 V + 8 V  16 V = 0, Vab = 16 V  12 V = 4 V
Vab + 12 V  18 V + 6 V  12 V = 0, Vab = 30 V  18 V = 12 V
19.
a.
ET = 8 V  16 V + 20 V = 12 V, I =
b.
20.
a.
12 V
= 1.17 A
10.3 
2V
= 173.91 mA
ET = 4 V + 18 V  12 V = 2 V, I 
11.5 
P = 8 mW = I2R, R =
I=
b.
21.
22.
8 mW
8 mW
= 2 k

2
(2 mA)2
I
E
20 V  E
= 2 mA (CW),

RT 3 k  2 k
16 V
V
12 V
= 8 mA, R =

= 1.5 k
2 k
I 8 mA
E
E  4 V  10 V E  14 V
= 8 mA (CCW)
I=


RT
2 k   1 .5 k 
3.5 k
E = 42 V
I=
a.
+10 V + 4 V  12 V  V = 0
V = 14 V  12 V = 2 V
b.
+30 V + 20 V  8 V  V = 0
V = 50 V  8 V = 42 V
c.
6 V  22 V  V1 + 36 V = 0
V = 36 V  28 V = 8 V
a.
I=
b.
V2 = IR = (1.5 A)(2 Ω) = 3 V
c.
Chapter 5
E = 10 V
12 V
= 1.5 A
8
60 V  12 V  V1  3 V = 0
V1 = 60 V  15 V = 45 V
35
23.
a.
b.
24.
a.
b.
25.
26.
a.
b.
d.
36
V1.8Ω = IR = (3 A)(1.8 Ω) = 5.4 V
24 V  V1  10 V  5.4 V = 0, V1 = 24 V  15.4 V = 8.6 V
V2.7Ω = IR = (3 A)(2.7 Ω) = 8.1 V
10 V  8.1 V  V2 = 0
V2 = 10 V  8.1 V = 1.9 V
+ 10 V  V1 + 6 V  2 V  3 V = 0, V1 = 11 V
+10 V  V2  3 V = 0, V2 = 7 V
(50 V)(2 )
1 V 50 V
, R2 =
= 100 

1V
2
R2
(100 V)(2 )
1 V 100 V
, R3 =
= 200 

2
R3
1V
c.
27.
+10 V  V2 = 0
V2 = 10 V
+10 V  6 V  V1 = 0
V1 = 4 V
+24 V  10 V  V1 = 0
V1 = 14 V
+10 V  V2 + 8 V = 0
V2 = 18 V
10 k
V3: V2 = 10 k:1 k = 10:1
V3: V1 = 10 k:100  = 100:1
RE
(10 k)(60 V)
= 54.05 V
V3 = 3 
0.1 k  1 k  10 k
RT
( R  R3 ) E (1 k  10 k)(60 V)
V = 2
= 59.46 V

11.1 k
RT
a.
V=
40 (30 V)
= 20 V
40   20 
b.
V=
(2 k  3 k)(40 V)
(5 k)(40 V)
= 20 V

4 k  1 k  2 k  3 k
10 k
c.
(1.5   0.6   0.9 )(0.72 V)
(3 )(0.72 V)
= 0.36 V

(2.5   1.5   0.6   0.9   0.5 )
6 k
Chapter 5
28.
a.
V1
20 V
(1.2 )(20 V)
, V1 =
= 12 V

1.2  2 
2
V2
20 V
(6.8 )(20 V)
, V2 =
= 68 V

6.8  2 
2
E = V1 + 20 V + V2 = 12 V + 20 V + 68 V = 100 V
b.
c.
d.
29.
120 V  V1  80 V = 0, V1 = 40 V
80 V  10 V  V3 = 0, V3 = 70 V
V
1000 V
68 (1000 V)
= 680 V
 2 , V2 =
100  68 
100 
1000 V V1
2 (1000 V)
, V1 =
= 20 V

100  2 
100 
E = V1 + V2 + 1000 V
= 20 V + 680 V + 1000 V
= 1700 V
V1 = 0 V
10 k(50 V  30 V)
V2 =
10 k  3.3 k  4.7 k
10 k(20 V)
= 11.11 V
=
18 k
Vx = E1  V3.3kΩ
3.3 k(20 V)
V3.3kΩ =
18 k
= 3.67 V
Vx = 50 V  3.67 V = 46.33 V
2V
V
2 k(2 V)
 2 , V2 =
=4V
1 k 2 k
1 k
2V
V
3 k(2 V)
 4 , V4 =
=6V
1 k 3 k
1 k
2V
= 2 mA
1 k
E = 2 V + 4 V + 12 V + 6 V = 24 V
I=
Chapter 5
37
30.
a.
b.
31.
a.
b.
32.
R (20 V)
2.2 k  1.8 k  R
4(4 kΩ + R) = 20R
16 kΩ + 4R = 20R
16R = 16 kΩ
16
R=
k = 1 kΩ
16
4V=
(6 M  R )(140 V)
6 M  R  3 M
100(9 MΩ + R) = 840 MΩ + 140R
900 MΩ + 100R = 840 MΩ + 140R
40R = 60 MΩ
60
R=
M = 1.5 MΩ
40
100 V =
8V
= 160 
50 mA
R (12 V) 160 (12 V)
, Rx = 80  in series with the bulb
Vbulb = 8 V = bulb

160   Rx
Rbulb  Rx
Rbulb =
VR = 12 V  8 V = 4 V, P =
V 2 (4 V) 2
= 0.2 W, 1/4 W okay

R
80 
VR1  VR2 = 72 V
1
VR  VR2 = 72 V
5 2
72 V
 1
VR2 1   = 72 V, VR2 
= 60 V
1.2
 5
VR
VR
60 V
72 V  60 V 12 V
R2 = 2 
= 15 k, R1 = 1 

= 3 k
I R2
4 mA
I R1
4 mA
4 mA
33.
RT = R1 + R2 + R3 = 2R3 + 7R3 + R3 = 10R3
R (60 V)
VR3 = 3
= 6 V, VR1 = 2VR3 = 2 (6 V) = 12 V, VR2  7VR3 = 7(6 V) = 42 V
10 R3
34.
a.
VR3  4VR2 = 4(3VR1 )  12VR1
E = VR1  3VR1  12VR1 RT = R1 + 3R1 + 12R1 = 16R1 =
R1 =
38
64 V
= 6.4 k
10 mA
6.4 k
= 400 , R2 = 3R1 = 1.2 k, R3 = 12R1 = 4.8 k
16
Chapter 5
35.
36.
6.4 M
64 V
= 6.4 M, R1 =
= 400 k, R2 = 1.2 M, R3 = 4.8 M
10 A
16
I1 10 mA
R  400 k

= 103 and 1 
= 103 also
I  10  A
R1
400 
b.
RT =
a.
Va = 12 V + 5 V = 17 V
Vb = 5 V + 16 V = 21 V
Vab = 17 V  21 V = 4 V
b.
Va = 14 V
Vb = 14 V + 6 V + 10 V = 30 V
Vab = 14 V  30 V = 16 V
c.
Va = 10 V + 3 V = 13 V
Vb = 8 V
Vab = 13 V  (8 V) = 21 V
a.
60 V + 20 V 80 V

= 0.8 A
18   82  100 
Va = 60 V  I(18 Ω) = 60 V  (0.8 A)(18 Ω) = 60 V  14.4 V = 45.6 V
I =

I =
100 V + 60 V
160 V

= 20 mA
2 k  2 k  2 k  2 k 8 k
Va = 10 V  I(2 kΩ) = 100 V  (20 mA)(2 kΩ) = 100 V  40 V = 60 V
37.
38.
47 V  20 V
27 V
=
= 3 mA (CCW)
2k +3k + 4k 9k
V2k = 6 V, V3k = 9 V, V4k = 12 V
I=
a.
Va = 20 V, Vb = 20 V + 6 V = 26 V, Vc = 20 V + 6 V + 9 V = 35 V
Vd = 12 V, Ve = 0 V
b.
Vab = 6 V, Vdc = 47 V, Vcb = 9 V
c.
Vac = 15 V, Vdb = 47 V + 9 V = 38 V
VR
4 V + 4 V 8V
12 V  4 V 8 V

= 1 A, R1 = 1 =
= 8 Ω,

8
8
I
1A
1A
VR
8V 4V 4V
R3 = 3 =
=4Ω

I
1A
1A
I R2 =
Chapter 5
39
39.
VR2 = 48 V  12 V = 36 V
VR2
36 V
= 2.25 k
I
16 mA
VR3 = 12 V  0 V = 12 V
R2 =
=
VR3
12 V
=
= 0.75 k
I
16 mA
VR4 = 20 V
R3 =
VR4
20 V
= 1.25 k
I
16 mA
VR1  E  VR2  VR3  VR4
R4 =
=
= 100 V  36 V  12 V  20 V = 32 V
VR
32 V
= 2 k
R1 = 1 =
I
16 mA
40.
41.
42.
40
a.
Va = 8 V + 14 V = +6 V, Vb = 14 V
Vc = +I(10 Ω)  6 V with
14 V + 6 V 20 V
I=

=1A
10   10  20 
Therefore, Vc = (1 A)(10 Ω)  6 V = 10 V  6 V = 4 V
Vd = 0 V
b.
Vab = Va  Vb = 6 V  14 V = 8 V
Vcb= Vc  Vb = 4 V  14 V = 10 V
Vcd = Vc  Vd = 4 V  0 V = 4 V
c.
Vad = Va  Vd = 6 V  0 V = 6 V
Vca = Vc  Va = 4 V  6 V = 2 V
V0 = 0 V, V4 = (2 kΩ)(6 mA) + 3 V = 12 V + 3 V = 15 V, V7 = 4 V
V10 = V1  V0 = 12 V  0 V = 12 V, V23 = V2  V3 = 4 V  (8 V) = 4 V + 8 V = 12 V
V30 = V3  V0 = 8 V  0 V = 8 V, V67 = V6  V7 = 4 V  4 V = 0 V
4 V + 8 V 12 V

=3A
V56 = V5  V6 = 3 V  4 V = 1 V, I =
4
4
V0 = 0 V, V03 = V0  V3 = 0 V  0 V = 0 V, V2 = (3 mA)(3.3 kΩ) = 9.9 V
V23 = V2  V3 = 9.9 V  0 V = 9.9 V, V12 = V1  V2 = 20 V  9.9 V = 10.1 V
 Ii   Io
Ii = 4 mA + 3 mA + 10 mA = 17 mA
Chapter 5
43.
44.
45.
a.
VL = ILRL = (2 A)(28 ) = 56 V
Vint = 60 V  56 V = 4 V
V
4V
=2
Rint = int 
I
2A
b.
VR =
60 V  56 V
VNL  VFL
 100% =
 100% = 7.14%
VFL
56 V
a.
VL =
3.3 (12 V)
39.6 V
= 11.85 V

3.3   43 m 3.343 
b.
VR =
12 V  11.85 V
VNL  VFL
 100% =
 100% = 1.27%
11.85 V
VFL
c.
Is = IL =
a.
I=
E
12 V
12 V
= 1.36 mA


RT 2 k  6.8 k 8.8 k
b.
I=
E
12 V
12 V
= 1.33 mA


RT 8.8 k  0.25 k 9.05 k
c.
not for most applications.
Chapter 5
11.85 V
= 3.59 A
3.3 
Ps = EIs = (12 V)(3.59 A) = 43.08 W
Pint = I2Rint = (3.59 A)2 43  = 0.554 W
41
Chapter 6
1.
a.
b.
c.
d.
e.
f.
g.
R2 and R3
E and R3
R2 and R3
R2 and R3
E, R1, R2, R3, and R4
E, R1, R2, and R3
E2, R2 and R3
2.
a.
b.
R3 and R4, R5 and R6
E and R1, R6 and R7
3.
a.
RT =
b.
RT =
c.
RT =
d.
e.
4.
42
(36 )(18 )
= 12 
36   18 
1
1

3
1
1
1
1  10 S  0.5  103 S  33.33  106 S


1 k 2 k 30 k
1
=
= 0.652 k
1.533  103 S
1
1
1


6
3
3
1
1
1
833.33  10 S  8.33  10 S  83.33  10 S 92.49  103 S


1.2  120 k 12 k
= 10.81 
18 k
= 6 k
3
(6 k)(6 k)
= 3 k
RT =
6 k  6 k
RT =
22 
10 
= 5.5 , RT  =
=5
4
2
(5.5 )(5 )
= 2.62 
RT =
5.5   5 
RT =
1
1

3
1
1
1
1000  10 S  1  10 3 S  0.001  103 S


1  1 k 1 M
1
=
= 0.99 
1001.001  10 3 S
f.
RT =
a.
RT =
1
1


3
1
1
1
1  10 S  0.833  103 S  3.333  10 3 S


1 k  1 .2 k  0 .3 k 
1
= 193.57 
=
5.166  10 3 S
Chapter 6
5.
1

1
1  10 S  0.833  10 S  0.455  10 3 S  1  103 S
b.
RT =
a.
RT = 3   6  = 2 
(2 )( R )
, R=8
RT = 1.6  =
2R
6 k
RT =
= 2 k
3
(2 k)( R)
, R = 18 k
RT = 1.8 k =
2 k  R
(20 k)( R )
RT = 5.08 k =
, R = 6.8 k
20 k  R
1
1

RT = 1.02  =
1
1
1
1
416.67  10 6 S +  147.06  10 6 S
 
2.4 k R 6.8 k
R
1
1.02 kΩ =
1
563.73  106 
R
1.020
k

575 × 103 +
=1
R
1.020 k
R=
= 2.4 kΩ
425  103
b.
c.
d.
e.
1
1
1
1



1 k 1.2 k 2.2 k 1 k
1
=
= 304.14 
3.288  103 S
RT = 6 kΩ =
3
3
R1
4
R1 = 24 kΩ
6.
a.
b.
1.2 k
about 1 k
c.
RT =
d.
e.
Chapter 6
1
1
1
1
1



1.2 k 22 k 220 k 2.2 M
1
=
6
6
833.333  10 S  45.455  10 S  4.545  10 6 S  0.455  10 6 S
1
= 1.131 k
=
883.788  10 6 S
(1.2 k)(22 k)
220 k, 2.2 M: RT =
= 1.138 k
1.2 k  22 k
RT reduced.
43
7.
1
1
1


= 1.18 
1
1
1
0.25 S + 0.50 S  0.10 S 0.85 S


4  2  10 
a.
RT =
b.

c.
RT = 3 Ω  6 Ω = 2 
8.
24   24  = 12 
1
1
1
1



RT R1 12  120 
1
0.1 S =
+ 0.08333 S + 0.00833 S
R1
1
+ 0.09167 S
0.1 S =
R1
1
= 0.1 S  0.09167 S = 0.00833 S
R1
1
= 120 
R1 =
0.00833 S
9.
a.
RT =
b.
VR1  VR2 = 36 V
c.
Is =
d.
Is = I1 + I2
6 A = 4.5 A + 1.5 A = 6 A (checks)
a.
I1 =
b.
RT =
c.
Is =
10.
44
(8 )(24 )
=6
8   24 
E 36 V
=6A

RT
6
VR 36 V
= 4.5 A
I1  1 
8
R1
VR
36 V
= 1.5 A
I2  2 
R2 24 
VR1
R1

VR 18 V
VR 18 V
18 V
= 6 A, I2 = 2 
= 2 A, I3 = 3 
= 0.5 A
3
9
R2
R3 36 
1
1

1
1
1

0
.
333
S
0
.
111
S  0.028 S


3  9  36 
1
= 2.12 
=
472  103 S
E
18 V
= 8.5 A

RT 2.12 
Chapter 6
11.
d.
Is = I1 + I2 + I3 = 6 A + 2 A + 0.5 A = 8.5 A
e.
they match
a.
I R1 
I R3 
12.
VR1
R1
VR3
R3

VR
24 V
24 V
= 2.4 mA, I R2  2 
= 20 mA,
10 k
R 2 1.2 k

24 V
= 3.53 mA
6.8 k
1
1


6
1
1
1
100  10 S  833.333  106 S  147.06  106 S


10 k 1.2 k 6.8 k
1
= 925.93 
=
1.08  103 S
b.
RT =
c.
Is =
d.
Is = I1 + I2 + I3 = 2.4 mA + 20 mA + 3.53 mA = 25.93 mA
e.
they match
a.
RT  900 
b.
RT =
c.
I3 the most, I4 the least
d.
I R1 
e.
Is =
f.
always greater
Chapter 6
E
24 V
= 25.92 mA

RT 925.93 
1
1
1
1
1



20 k 10 k 1 k 91 k
1
=
6
6
50  10 S  100  10 S  1  103 S  10.99  106 S
1
= 862.07 , very close
=
1.16  103 S
VR1
VR
60 V
60 V
= 3.0 mA, I R2  2 
= 6 mA
R1 20 k
R2 10 k
VR
VR
60 V
60 V
= 60.0 mA, I R4  4 
= 0.659 mA
I R3  3 
R3 1 k
R4 91 k

E
60 V
= 69.6 mA

RT 862.07 k
Is = 3 mA + 6 mA + 60 mA + 0.659 mA = 69.66 mA (checks)
45
13.
14.
RT = 6 Ω =
b.
P = 81 W =
a.
P=
V 2 E2

and E =
R
R
b.
R2 =
E 20 V
= 10 Ω

I2
2A
c.
I1 =
V1 E 20 V
=2A


R1 R1 10 
d.
Is = I1 + I2 + I3 = 2 A + 2 A +
e.
Ps = EIs = (20 V)(9 A) = 180 W
f.
g.
15.
(18 )( R2 )
18   R2
108 Ω + 6R2 = 18R2
12R2 = 108 Ω
108 
=9Ω
R2 =
12
a.
V 2 E2 E2


R
R 9
and E2 = (9)(81)
or E = 729 = 27 V
PR  (100 W)(4 )  400 = 20 V
20 V
=4A+5A=9A
4
E 2 (20 V) 2 400 W
E 2 (20 V) 2
= 40 W, PR2 
= 40 W,


PR1 

10
R2  20 V 
R1
10 


2A 
Ps = P1 + P2 + P3
180 W = 40 W + 40 W + 100 W = 180 W (checks)
(20 )(10.8 A)
=9A
20   4 
E = VR3  I 3 R3  (9 A)(4 ) = 36 V
I3 =
I R1 = 12.3 A  10.8 A = 1.5 A
R1 =
16.
46
VR1
I R1

36 V
= 24 
1.5 A
a.
V = 48 V
b.
I2 =
48 V
= 2.67 mA
18 k
Chapter 6
17.
c.
Is =
48 V 48 V

 I 2 = 16 mA + 4 mA + 2.67 mA = 22.67 mA
3 k 12 k
d.
P=
V 2 E 2 (48 V)2


= 192 mW
R
R
12 k
a.
I R2  = 4 A  1 A = 3 A, R2 =
b.
R3 =
c.
I1 
18.

19.
a.
VR3
I2

E 12 V
=4Ω

I2 3 A
E 12 V
= 12 Ω

I3 1 A
12 V
= 6 A, Is = I1 + 4 A = 6 A + 4 A = 10 A
2
1
1


6
1
1
1
1000  10 S  212.77  106 S  100  106 S


1 k 4.7 k 10 k
1
=
= 761.61 
1.313  103 S
VR 60 V
VR
60 V
= 60 mA, I R2  2 
= 12.77 mA
I R1  1 
R1 1 k
R2 4.7 k
RT =
I R3 
b.
I3

VR2
VR3
R3

60 V
= 6 mA
10 k
PR1  VR1  I R1 = (60 V)(60 mA) = 3.6 W
PR2  VR2  I R2 = (60 V)(12.77 mA) = 766.2 mW
PR3  VR3  I R3 = (60 V)(6 mA) = 360 W
d.
e.
E
60 V
= 78.78 mA

RT 761.61 
Ps = EsIs = (60 V)(78.78 mA) = 4.73 W
Ps = 4.73 W = 3.6 W + 766.2 mW + 360 mW = 4.73 W (checks)
R1 = the smallest parallel resistor
a.
Ibulb =
c.
20.
b.
c.
Chapter 6
Is =
E
120 V
= 66.667 mA

Rbulb 1.8 k
R 1.8 k
= 225 
RT =

N
8
E 120 V
= 0.533 A
Is =

RT 225 
47
21.
V 2 (120 V)2
=8W

R
1.8 k
d.
P=
e.
Ps = 8(8 W) = 64 W
f.
none, Is drops by 66.667 mA
Network redrawn:
RT = 3.33 Ω  7.5 Ω = 2.31 Ω
E 2 (60 V)2
= 1.56 kW
Ps =

RT
2.31 
22.
48
a.
5 × 60 W = 300 W
300 W
= 2.5 A
Ibulbs =
120 V
1200 W
Imicro =
= 10 A
120 V
320 W
ITV =
= 2.67 A
120 V
25 W
= 208.33 mA
IDVD =
120 V
b.
Is =  I = 2.5 A + 10 A + 2.67 A + 208.33 mA = 15.38 A
No
c.
RT =
d.
Ps = E Is = (120 V)(15.38 A) = 1,845.60 W
e.
Ps = 1845.60 W = 300 W + 1200 W + 320 W + 25 W = 1845 W (checks)
E
120 V
= 7.8 Ω

I s 15.38 A
Chapter 6
23.
a.
b.
c.
8   12  = 4.8 , 4.8   4  = 2.182 
24 V  8 V
= 14.67 A
I1 =
2.182 
V 2 (24 V  8 V) 2
P4 =
= 256 W

R
4
I2 = I1 = 14.67 A
24.
Is = 8 mA + 6 mA = 14 mA
I2 = 6 mA  2 mA = 4 mA
25.
a.
 Ii   Io
2A+3A+9A=6A+I
14 A = 6 A + I
I = 14 A  6 A = 8 A
b.
 Ii   Io
8 mA = 2 mA + I1
I1 = 8 mA  2 m A = 6 mA
 Ii   Io
I1 + 9 mA = I2
I2 = 6 mA + 9 mA = 15 mA
 Ii   Io
I2 = 10 mA + I3
I3 = 15 mA  10 mA = 5 mA
a.
 Ii   Io
8 A = 3 A + I2
I2 = 8 A  3 A = 5 A, I3 = 3 A
 Ii   Io
I2 + I3 = I4
I4 = 5 A + 3 A = 8 A
b.
 Ii   Io
Is = 36 mA + 4 mA = 40 mA
 Ii   Io
36 mA = I3 + 20 mA
I3 = 36 mA  20 mA = 16 mA
 Ii   Io
4 mA + 20 mA = I4
I4 = 24 mA
I5 = Is = 40 mA
26.
Chapter 6
49
27.
I R2 = 5 mA  2 mA = 3 mA
E = VR2 = (3 mA)(4 k) = 12 V
R1 =
R3 =
28.
VR1
I R1
VR3
I R3

12 V
12 V

= 3 k
(9 mA  5 mA) 4 mA

12 V
= 6 k
2 mA
RT =
12 V
E
= 1.33 k

I T 9 mA
a.
R1 =
b.
E = I1R1 = (2 A)(6 ) = 12 V
E 12 V
= 1.33 A
I2 =

R2
9
E 10 V
=5

I1 2 A
I2 = I  I1 = 3 A  2 A = 1 A
E 10 V
R=
= 10 

I2
1A
P 12 W
=1A

V 12 V
E 12 V
= 12 
R3 =

I3
1A
I = I1 + I2 + I3 = 2 A + 1.33A + 1 A = 4.33 A
I3 =
29.
50
a.
64 V
= 64 mA
1 k
64 V
I3 =
= 16 mA
4 k
Is = I1 + I2 + I3
I2 = Is  I1  I3 = 100 mA  64 mA  16 mA = 20 mA
64 V
E
= 3.2 k
R=

I 2 20 mA
I = I2 + I3 = 20 mA + 16 mA = 36 mA
I1 =
Chapter 6
b.
V12
 V1  PR1  (30 W)(30 ) = 30 V
R1
E = V1 = 30 V
E 30 V
=1A
I1 =

R1 30 
Because R3 = R2, I3 = I2 , and Is = I1 + I2 + I3 = I1 + 2I2
2 A = 1 A + 2I2
1
I2 = (1 A) = 0.5 A
2
I3 = 0.5 A
E 30 V
R2 = R3 =
= 60 

I 2 0.5 A
P=
PR2  I 22 R2 = (0.5 A)2  60  = 15 W
6
1
1
I1  I1 = (9 A) = 4.5 A
12 
2
2
6
I3 =
I1  3I1 = 3(9 A) = 27 A
2
6
1
1
I4 =
I1  I1 = (9 A) = 3 A
18 
3
3
IT = I1 + I2 + I3 + I4 = 9 A + 4.5 A + 27 A + 3 A = 43.5 A
30.
I2 =
31.
a.
b.
32.
a.
8 k(20 mA )
= 16 mA
2 k  8 k
I2 = 20 mA  16 mA = 4 mA
I1 =
1 k( IT )
1 k( IT )

1 k  2.4 k
3.4 k
3.4 k(2.5 A)
and IT =
= 8.5 A
1 k
I1 = IT  2.5 A = 8.5 A  2.5 A = 6 A
I2.4kΩ = 2.5 A =
1
1
=

3
1
1
1
250  10 S  125  103 S  83.333  10 3 S


4  8  12 
1
=
= 2.18 
458.333  103
R
2.18 
I1 =
(6 A) = 3.27 A
Ix = T I,
4
Rx
RT =
2.18 
(6 A) = 1.64 A
8
2.18 
I3 =
(6 A) = 1.09 A
12 
I4 = 6 A
I2 =
Chapter 6
51
b.
33.
9
(10 A) = 9 A
10
a.
I1 
b.
I1/I2 = 10 /1  = 10,
c.
I1/I3 = 1 k/1  = 1000, I3 = I1/1000 = 9 A/1000  9 mA
d.
I1/I4 = 100 k/1  = 100,000, I4 = I1/100,000 = 9 A/100,000  90 A
e.
very little effect, 1/100,000
1
RT =
1
1
1
1



1  10  1 k 100 k
1
=
1 S  0.1 S  1  10 3 S  10  10 6 S
1
=
= 0.91 
1.10 S
R
0.91 
Ix = T I , I1 =
(10 A) = 9.1 A excellent (9 A)
1
Rx
f.
I2 =
I1 9 A

 0.9 A
10 10
g.
I2 =
0.91 
(10 A) = 0.91 A excellent (0.9 A)
10 
h.
I3 =
0.91 
(10 A) = 9.1 mA excellent (9 mA)
1 k
i.
34.
4 Ω  4 Ω = 2 Ω
20 (8 A)
20 (8 A)

= 5.33 A
I2 =
20   2   8 
30 
I
5.33 A
I1 = 2 
= 2.67 A
2
2
I3 = 8 A  I2 = 8 A  5.33 A = 2.67 A
I4 = 8 A
a.
0.91 
(10 A) = 91 A excellent (90 A)
100 k
3 I
(39 )(1 A)
CDR: I36 =
= 1 A, I =
= 13 A = I2
3   36 
3
I4 =
I1 = I  1 A = 13 A  1 A = 12 A
b.
52
I3 = I = 24 mA, V12kΩ = IR = (4 mA)(12 kΩ) = 48 V
V 48 V
= 12 mA
I2 = 
R 4 k
I1 = I 4 mA  I2
= 24 mA  4 mA  12 mA
= 8 mA
Chapter 6
35.
a.
b.
R = 3(2 k) = 6 k
6 k(32 mA)
= 24 mA
I1 =
6 k  2 k
I
24 mA
I2 = 1 
= 8 mA
3
3
36.
84 mA = I1 + I2 + I3 = I1 + 2I1 + 2I2 = I1 + 2I1 + 2(2I1)
84 mA = I1 + 2I1 + 4I1 = 7I1
84 mA
and I1 =
= 12 mA
7
I2 = 2I1 = 2(12 mA) = 24 mA
I3 = 2I2 = 2(24 mA) = 48 mA
VR
24 V
= 2 k
R1 = 1 
I1 12 mA
VR
24 V
= 1 k
R2 = 2 
24 mA
I2
VR
24 V
= 0.5 k
R3 = 3 
I 3 48 mA
37.
a.
38.
39.
b.
c.
PL = VLIL
72 W = 12 V  IL
72 W
=6A
IL =
12 V
I
6A
I1 = I2 = L 
=3A
2
2
Psource = EI = (12 V)(3 A) = 36 W
Ps1  Ps2 = 36 W + 36 W = 72 W (the same)
d.
Idrain = 6 A (twice as much)
RT = 8   56  = 7 
E 12 V
= 1.71 A
I2 = I3 =

RT
7
1
1
I1 = I 2  (1.71 A) = 0.86 A
2
2
16 V
= 2 A, I = 5 A  2 A = 3 A
8
V
16 V
=2
IR = 5 A + 3 A = 8 A, R = R 
IR
8A
I8  =
Chapter 6
53
40.
c.
E
12 V
12 V
= 1.188 mA


RT 0.1 k  10 k 10.1 k
VL = IsRL = (1.19 mA)(10 k) = 11.90 V
12 V
= 120 mA
Is =
100 
VL = E = 12 V
a.
VL =
a.
b.
41.
b.
c.
42.
43.
4.7 k(9 V)
42.3V

= 6.13 V
4.7 k  2.2 k
6.9
VL = E = 9 V
VL = E = 9 V
b.
c.
20 V
= 5 A, I2 = 0 A
4
V1 = 0 V, V2 = 20 V
Is = I1 = 5 A
a.
V2 =
b.
RT = 11 M  22 k = 21.956 k
21.956 k(20 V)
= 16.47 V (very close to ideal)
V2 =
21.956 k  4.7 k
c.
Rm = 20 V[20,000 /V] = 400 k
RT = 400 k  22 k = 20.853 k
20.853 k(20 V)
= 16.32 V (still very close to ideal)
V2 =
20.853 k  4.7 k
d:
a.
V2 =
b.
RT = 200 k  11 M = 196.429 k
(196.429 k)(20 V)
= 13.25 V (very close to ideal)
V2 =
196.429 k  100 k
c.
Rm = 400 k
RT = 400 k  200 k = 133.333 k
(133.333 k)(20 V)
= 11.43 V (a 1.824 V drop from Rint = 11 M level)
V2 =
133.333 k  100 k
a.
e.
54
Is =
I1 =
22 k(20 V)
= 16.48 V
22 k  4.7 k
200 k(20 V)
= 13.33 V
200 k  100 k
DMM level of 11 M not a problem for most situations
VOM level of 400 k can be a problem for some situations.
Chapter 6
44.
a.
Vab = 20 V
b.
Vab =
c.
Rm = 200 V[20,000 /V] = 4 M
4 M(20 V)
= 16.0 V (significant drop from ideal)
Vab =
4 M  1 M
Rm = 20 V[20,000 /V] = 400 k
400 k(20 V)
= 5.71 V (significant error)
Vab =
400 k  1 M
11 M(20 V)
= 18.33 V
11 M  1 M
45.
not operating properly, 6 k not connected at both ends
6V
= 1.71 k
RT =
3.5 mA
RT = 3 k  4 k = 1.71 k
46.
Vab = E + I4 k  R4 k
12 V  4 V
8V

= 1.6 mA
I4 k =
1 k  4 k 5 k
Vab = 4 V + (1.6 mA)(4 k) = 4 V + 6.4 V = 10.4 V
4 V supply connected in reverse so that
12 V  4 V 16 V

= 3.2 mA
I=
1 k  4 k 5 k
and Vab = 12 V  (3.2 mA)(1 k) = 12 V  3.2 V = 8.8 V obtained
Chapter 6
55
Chapter 7
1.
a.
b.
c.
d.
e.
f.
2.
a.
b.
c.
d.
R1, R2,. and E are in series; R3, R4 and R5 are in parallel
E and R1 are in series; R2, R3 and R4 are in parallel.
E and R1 are in series; R2, R3 and R4 are in parallel.
E1 and R1 are in series; E2 and R4 in parallel.
E and R1 are in series, R2 and R3 are in parallel.
E, R1, R4 and R6 are in parallel; R2 and R5 are in parallel.
RT = 4  + 10   (4  + 4 ) + 4  = 4 Ω + 10   8  + 4 
= 4  + 4.44  + 4  = 12.44 
10 
= 10  + 5  = 15 
RT = 10  +
2
4
+ 10  = 2  + 10  = 12 
RT =
2
RT = 10 
3.
2.2 k  10 k  = 1.8 k
RT = 2 × 1.8 kΩ = 3.6 kΩ
RT = 1 Ω  (1 Ω + 1 Ω + RT) = 1 Ω  (2 Ω + RT)
2   RT
2   RT
=

1   2   RT 3   RT
RT(3 Ω + RT) = 2 Ω + RT
3RT + RT2 = 2 Ω + RT
4.
RT2 + 2RT  2 Ω = 0
2  (2) 2  4(1)(2)
2
2  4  8 2  12 2  3.464
=


2
2
2
RT = 1  1.732 = 0.732 Ω or 2.732 Ω
Since RT < 1 Ω and positive choose RT = 0.732 Ω
RT =
56
Chapter 7
5.
R 

RT = 7.2 kΩ = R1   R1  1  = R1  1.5R1

2
( R1 )(1.5R1 ) 1.5 R12 1.5 R1


2.5 R1
2.5
R1  1.5 R1
2.5(7.2 k)
1.2 kΩ
and R1 =
1.5
so that 7.2 kΩ =
6.
a.
b.
c.
d.
e.
f.
g.
7.
a.
b.
c.
8.
yes
I2 = Is  I1 = 10 A  4 A = 6 A
yes
V3 = E  V2 = 14 V  8 V = 6 V
RT = 4   2  = 1.33  , RT = 4   6  = 2.4 
RT = RT  RT = 1.33  + 2.4  = 3.73 
20 
RT  RT =
= 10 , RT = RT  RT = 10  + 10  = 20 
2
E
20 V
=1A
Is =

RT 20 
Ps = EIs = Pabsorbed = (20 V)(1 A) = 20 W
RT = R1  R2 = 10   15  = 6 
RT = RT  (R3 + R4) = 6   (10  + 2 ) = 6   12  = 4 
E 36 V
E
36 V
= 9 A, I1 =
=
=6A
Is =

RT
4
RT
6
E
36 V
36 V
=3A
I2 =


R3  R4 10   2  12 
I1 = Is  I2 = 6 A  3 A = 3 A
Va = I2R4 = (3 A)(2 ) = 6 V
Redrawn:
a.
b.
Va = 32 V
8 Ω  24 Ω = 6 Ω
6 (32 V)
= 10.67 V
Vb =
6   12 
32 V
32 V

= 1.78 A
12 +6  18 
RT = 72 Ω  18 Ω  18 Ω = 8.12 Ω
I1 =
9Ω
Chapter 7
57
Is =
9.
a.
b.
E
32 V
= 3.94 A

RT 8.12 
Va = 36 V, Vb = 60 V Vc =
5 k(60 V)
= 20 V
5 k  10 k
 60 V  36 V
I1 
= 24 mA,
1 k
60 V
60 V
I8kΩ = 8 k = 7.5 mA, I10kΩ =
= 4 mA
15 k
24 mA 
 I  24 mA + 7.5 mA = 31.5 mA
 I 2  31.5 mA + 4 mA = 35.5 mA
10.
11.
a.
RT = 1.2 k + 6.8 k = 8 k, RT = 2 k  RT = 2 k  8 k = 1.6 k
RT  = RT + 2.4 k = 1.6 k + 2.4 k = 4 k
RT = 1 k  RT  = 1 k  4 k = 0.8 k
b.
Is =
E
48 V
= 60 mA

RT 0.8 k
c.
V=
(1.6 k)(48 V)
RTE
= 19.2 V

RT  2.4 k  1.6 k  2.4 k
RT = 2 R  2 R (R + R) = 2 R  2R  2 R =
2R
3
E 120 V

= 15 Ω
I
8A
2R
3
15 Ω =
and R = (15 ) = 22.5 Ω
3
2
2 R = 45 Ω
RT =
12.
58
a.
RT = (R1  R2  R3)  (R6 + R4  R5)
= (12 k  12 k  3 k)  (10.4 k + 9 k  6 k)
= (6 k  3 k)  (10.4 k + 3.6 k)
= 2 k  14 k = 1.75 k
E
28 V
E
28 V
= 16 mA, I2 =
= 2.33 mA
Is =


RT 1.75 k
R2 12 k
R = R1  R2  R3 = 2 k
R = R6 + R4  R5 = 14 k
Chapter 7
I6 =
Chapter 7
2 k(16 mA)
R( I s )
= 2 mA

R  R 2 k  14 k
59
b.
V1 = E = 28 V
R = R4  R5 = 6 k  9 k = 3.6 k
V5 = I6 R = (2 mA)(3.6 k) = 7.2 V
VR23
c.
P=
13.
a.
I1 
14.
I1 =
15.
a.
60
(28 V)2
= 261.33 mW
3 k
24 V
= 6 A; VR2  24 V  8 V = 16 V, I 2  VR2 / R2 = 16 V/2 Ω = 8 A
4
8V
I 3 
= 0.8 A, I = I1 + I2 = 6 A + 8 A = 14 A
10 
20 V
= 425.5 mA
47 
14 V
14 V
I2 =

= 139.35 mA
160   270  100.47 
b.
16.
R3

a.
R = R4 + R5 = 14 Ω + 6 Ω = 20 Ω
R = R2  R = 20 Ω  20 Ω = 10 Ω
R = R + R1 = 10 Ω + 10 Ω = 20 Ω
RT = R3  R = 5 Ω  20 Ω = 4 Ω
E 20 V
=5A
Is =
=
RT 4 
20 V
20 V
20 V
=1A
I1 =
=
=
R1  R 10  + 10  20 
20 V
=4A
I3 =
5
I
1A
I4 = 1 = (since R = R2) =
= 0.5 A
2
2
Va = I3R3  I4R5 = (4 A)(5 Ω)  (0.5 A)(6 Ω) = 20 V  3 V = 17 V
I 
Vbc =  1  R2 = (0.5 A)(20 Ω) = 10 V
2
E
20 V

R1  R4  ( R2  R3  R5 ) 3   3   (3   6   6 )
20 V
20 V
20 V
=
=
=
3  + 3   (3  + 3 ) 3  + 3   6  3  + 2 
=4A
I1 =
Chapter 7
17.
R4 ( I1 )
3 (4 A)

R4  R2  R3  R5 3   3   6   6 
12 A
= 1.33 A
=
6+3
I
I3 = 2 = 0.67 A
2
b.
CDR: I2 =
c.
I4 = I1  I2 = 4 A  1.33 A = 2.67 A
Va = I4R4 = (2.67 A)(3 Ω) = 8 V
Vb = I3R3 = (0.67 A)(6 Ω) = 4 V
a.
IE =
b.
IB =
c.
VB = VBE + VE = 2.7 V
VC = VCC  ICRC = 8 V  (2 mA)(2.2 kΩ) = 8 V  4.4 V = 3.6 V
d.
VCE = VC  VE = 3.6 V  2 V = 1.6 V
VE
2V
= 2 mA

RE 1 k
IC = IE = 2 mA
VCC  (VBE  VE ) 8 V  (0.7 V + 2 V)
=
220 k 
RB
RB
8 V  2.7 V
5.3 V
=
= 24 μA
=
220 k
220 k
VRB
=
VBC = VB  VC = 2.7 V  3.6 V = 0.9 V
18.
19.
22 V
22 V

=1A
4   18  22 
a.
I=
b.
22 V + Vi  22 V = 0, V1 = 44 V
a.
All resistors in parallel (between terminals a & b)
RT = 16 Ω  16 Ω  8 Ω  4 Ω  32 Ω
8 Ω  8 Ω  4 Ω  32 Ω
4 Ω  4 Ω  32 Ω
2 Ω  32 Ω = 1.88 
Chapter 7
61
b.
All in parallel. Therefore, V1 = V4 = E = 32 V
c.
I3 = V3/R3 = 32 V/4 Ω = 8 A 
d.
12 V
= 1.2 mA
10 k
V ab = Va  Vb = 12 V  (18 V) = 30 V
20.
I=
21.
a.
b.
22.
Is = I1 + I2 + I3 + I4 + I5
32 V 32 V 32 V 32 V 32 V
+
+
+
+
=
16  8 
4  32  16 
=2A+4A+8A+1A+2A
= 17 A
E 32 V
= 1.88 Ω as above
RT =
=
I s 17 A
a.
Va = 6 V, Vb = 20 V
Vab = Va  Vb = (6 V)  (20 V) = 6 V + 20 V = +14 V
20 V
=4A
5
V
14 V
I 2   ab 
=7A
2 2
6V
I 3 
=2A
3
I3Ω = I2Ω + I 6V , I6V = I3Ω  I2Ω = 2 A  7 A = 5 A
I + I6V = I5Ω, I = I5Ω  I6V = 4 A  (5A) = 9 A
I 5  
Applying Kirchoff's voltage law in the CCW direction in the upper "window":
+18 V + 20 V  V8Ω = 0
V8Ω = 38 V
38 V
= 4.75 A
8
18 V
18 V
I3Ω =
=
=2A
3 + 6 9
I8Ω =
KCL: I18V = 4.75 A + 2 A = 6.75 A
b.
62
V = (I3Ω)(6 Ω) + 20 V = (2 A)(6 Ω) + 20 V = 12 V + 20 V = 32 V
Chapter 7
23.
I2R2 = I3R3 and I2 =
I1 = I2 + I3 =
I 3 R3 2 R3 R3
(since the voltage across parallel elements is the same)


20 10
R2
R3
+2
10
R

KVL: 120 = I112 + I3R3 =  3  2  12 + 2R3
 10

and 120 = 1.2R3 + 24 + 2R3
3.2R3 = 96 
96 
= 30 
R3 =
3.2
24.
Assuming Is = 1 A, the current Is will divide as determined by the load appearing in each
branch. Since balanced Is will split equally between all three branches.
10
1 
V1 =  A  (10 ) = V
3
3 
10
1 
V2 =  A  (10 ) = V
6
6 
10
1 
V3 =  A  (10 ) = V
3
3 
10
10
10
E = V1 + V2 + V3 =
V + V + V = 8.33 V
3
6
3
E 8.33 V
= 8.33 
RT = =
I
1A
25.
36 kΩ  6 kΩ  12 kΩ = 3.6 kΩ
3.6 k (45 V)
= 16.88 V  27 V. Therefore, not operating properly!
V=
3.6 k  + 6 k 
6 kΩ resistor "open"
R (45V)
9 k(45 V)
= 27 V
R = 12 k  36 k = 9 k, V =

R  6 k 9 k  6 k
Chapter 7
63
26.
a.
RT = R5  (R6 + R7) = 6   3  = 2 
RT = R3  (R4 + RT) = 4   (2  + 2 ) = 2 
RT = R1 + R2 + RT = 3  + 5  + 2  = 10 
240 V
= 24 A
I=
10 
b.
I4 =
c.
d.
27.
28.
a.
4 ( I )
4 (24 A)
= 12 A

44
8
6 (12 A) 72 A
=8A
I7 =

6  3
9
V3 = I3R3 = (I  I4)R3 = (24 A  12 A)4 Ω = 48 V
V5 = I5R5 = (I4  I7)R5 = (4 A)6 Ω = 24 V
V7 = I7R7 = (8 A)2 Ω = 16 V
P = I 72 R7 = (8 A)22 Ω = 128 W
P = EI = (240 V)(24 A) = 5760 W
RT = R4  (R6 + R7 + R8) = 2 Ω  7 Ω = 1.56 Ω
RT = R2  (R3 + R5 + RT) = 2 Ω  (4 Ω + 1 Ω + 1.56 Ω) = 1.53 Ω
RT = R1 + RT = 4 Ω + 1.53 Ω = 5.53 Ω
b.
I = 40 V/5.53 Ω = 7.23 A
c.
I3 =
2 ( I )
2 (7.23 A)

= 1.69 A
2   6.56  2   6.56 
2 (1.69 A)
= 0.375 mA
I7 =
2+7
PR7  I 2 R = (0.375 A)2 2 Ω = 0.281 W
Network redrawn:
24 V
=3A
8
P6Ω = I2R = (3 A)2  6 Ω = 54 W
I8Ω = I6Ω =
64
Chapter 7
29.
a.
b.
30.
R10 + R11  R12 = 1 Ω + 2 Ω  2 Ω = 2 Ω
R4  (R5 + R6) = 10 Ω  10 Ω = 5 Ω
R1 + R2  (R3 + 5 Ω) = 3 Ω + 6 Ω  6 Ω = 6 Ω
RT = 2 Ω  3 Ω  6 Ω = 2 Ω  2 Ω = 1 Ω
I = 12 V/1 Ω = 12 A
I1 = 12 V/6 Ω = 2 A
6 (2 A)
=1A
I3 =
6+6
1A
I4 =
= 0.5 A
2
c.
I6 = I4 = 0.5 A
12 A
=6A
2
d.
I10 =
a.
E = (40 mA)(1.6 k) = 64 V
b.
RL2 =
c.
I R1 = 72 mA  40 mA = 32 mA
48 V
= 4 k
12 mA
24 V
RL3 =
= 3 k
8 mA
I R2 = 32 mA  12 mA = 20 mA
I R3 = 20 mA  8 mA = 12 mA
R1 =
R2 =
R3 =
31.
VR1
I R1
VR2
I R2
VR3
I R3
=
64 V  48 V 16 V
=
= 0.5 k
32 mA
32 mA
=
48 V  24 V
24 V
=
= 1.2 k
20 mA
20 mA
=
24 V
= 2 k
12 mA
I R1 = 40 mA
I R2 = 40 mA  10 mA = 30 mA
I R3 = 30 mA  20 mA = 10 mA
I R5 = 40 mA
I R4 = 40 mA  4 mA = 36 mA
R1 =
R2 =
Chapter 7
VR1
I R1
VR2
I R2
=
120 V  100 V
20 V
=
= 0.5 k
40 mA
40 mA
=
100 V  40 V
60 V
=
= 2 k
30 mA
30 mA
65
R3 =
R4 =
R5 =
VR3
I R3
VR4
I R4
VR5
I R5
=
40 V
= 4 k
10 mA
=
36 V
= 1 k
36 mA
=
60 V  36 V
24 V
=
= 0.6 k
40 mA
40 mA
P1 = I12 R1 = (40 mA)20.5 k = 0.8 W (1 watt resistor)
P2 = I 22 R2 = (30 mA)22 k = 1.8 W (2 watt resistor)
P3 = I 32 R3 = (10 mA)24 k = 0.4 W (1/2 watt or 1 watt resistor)
P4 = I 42 R4 = (36 mA)21 k = 1.3 W (2 watt resistor)
P5 = I 52 R5 = (40 mA)20.6 k = 0.96 W (1 watt resistor)
All power levels less than 2 W. Four less than 1 W.
32.
80 V
= 400 Ω  390 Ω
200 mA
40 V
R2 =
= 266.67 Ω  270 Ω
150 mA
R1 =
33.
a.
yes, RL  Rmax (potentiometer)
b.
VDR: VR2 = 3 V =
c.
VR1 = E  VL = 12 V  3 V = 9 V (Chose VR1 rather than VR2  RL since numerator of VDR
R2 (12 V) R2 (12 V)
=
R1  R2
1k
3 V(1 k )
= 0.25 k = 250 
R2 =
12 V
R1 = 1 k  0.25 k = 0.75 k = 750 
equation "cleaner")
R1 (12 V)
R1  ( R2  RL )
9R1 + 9(R2  RL) = 12R1
R1  3( R2  RL ) 
 2 eq. 2 unk( RL = 10 k)
R1  R2  1 k  
VR1 = 9 V =
R1 =
66
3R2 RL
3R2 10 k

R2  RL
R2  10 k
and R1(R2 + 10 k) = 30 k R2
Chapter 7
R1R2 + 10 k R1 = 30 k R2
R1 + R2 = 1 k: (1 k  R2)R2 + 10 k (1 k  R2) = 30 k R2
R22 + 39 k R2  10 k2 = 0
R2 = 0.255 k, 39.255 k
R2 = 255 
R1 = 1 k  R2 = 745 
34.
Vab =
b.
80   1 k = 74.07 
20   10 k = 19.96 
74.07 (40 V)
= 31.51 V
Vab =
74.07  + 19.96 
Vbc = 40 V  31.51 V = 8.49 V
c.
P=
d.
35.
36.
37.
38.
80 (40 V)
= 32 V
100 
Vbc = 40 V  32 V = 8 V
a.
(31.51 V ) 2 (8.49 V ) 2
+
= 12.411 W + 3.604 W = 16.02 W
80 
20 
(32 V ) 2 (8 V ) 2
+
= 12.8 W + 3.2 W = 16 W
80 
20 
The applied loads dissipate less than 20 mW of power.
P=
a.
ICS = 1 mA
b.
Rshunt =
Rm I CS
(100 )(1 mA) 0.1

Ω = 5 mΩ
=
20 A  1 mA
20
I max  I CS
(1 k )(50  A)
2Ω
25 mA  0.05 mA
(1 k )(50  A)
=1Ω
50 mA: Rshunt =
50 mA  0.05 mA
100 mA: Rshunt  0.5 Ω
25 mA: Rshunt =
Vmax  VVS 15 V  (50  A)(1 k)
= 300 kΩ
=
50  A
I CS
a.
Rs =
b.
Ω/V = 1/ICS = 1/50 μA = 20,000
5 V  (1 mA)(1000 )
= 4 kΩ
1 mA
50 V  1 V
50 V: Rs =
= 49 kΩ
1 mA
500 V  1 V
= 499 kΩ
500 V: Rs =
1 mA
5 V: Rs =
Chapter 7
67
39.
40.
10 MΩ = (0.5 V)(Ω/V)  Ω/V = 20  106
1
= 0.05 μA
ICS = 1/(Ω/V) =
20  106
a.
Rs =
b.
xIm =
2 k
E
zero adjust
3V
 1 kΩ 
= 28 kΩ
 Rm 
=
2
100  A
2
Im
Runk =
E
Rseries
+ Rm +
zero adjust
+ Runk
2
zero adjust 
E

  Rseries  Rm +

2
xI m


3V
30  103
 30 kΩ 
 30  103
x100  A
x
3
1
1
x = , Runk = 10 kΩ; x = , Runk = 30 kΩ; x = , Runk = 90 kΩ
4
2
4
=
41.

40.
a.
Carefully redrawing the network will reveal that all three resistors are in parallel
R 12 
=4Ω

and RT =
N
3
b.
Again, all three resistors are in parallel and RT =
a.
Network redrawn:
42.
R 18 
=6Ω

N
3
Rohmmeter = 1.2 kΩ  (3.1 kΩ + 1.2 kΩ + 1.65 kΩ)
= 1.2 kΩ  5.95 kΩ
= 1 kΩ
b.
All three resistors are in parallel
Rohmmeter =
68
R 18 

=6Ω
N
3
Chapter 7
Chapter 8
1.
2.
3.
8 (6 A)
= 4.8 A
8 2 
I2 = 6 A  I1 = 6 A  4.8 A = 1.2 A
a.
I1 =
b.
Vs = I1R1 = (4.8 A)(2 Ω) = 9.6 V
a.
I1 = I2 = 20 mA
b.
V2 = I2R2 = (20 mA)(3.3 kΩ) = 66 V
Vs = IRT = (20 mA)((1.2 kΩ + 3.3 kΩ) = 20 mA(4.5 kΩ) = 90 V
E + VR1  Vs = 0, VR1 = (8 mA)(2.7 kΩ) = 21.6 V
Vs = E + VR1 = 10 V + 21.6 V = 31.6 V 
4.
a.
b.
c.
Vs = E = 24 V
E
24 V
24 V
=6A
I2 =


R1  R2 1   3 
4
I + Is = I2, Is = I2  I = 6 A  2 A = 4 A
5.
V1 = V2 = Vs = IRT = 0.6 A[6   24   24 ] = 0.6 A[6   12 ] = 2.4 V
V
2.4 V
= 0.1 A
I2 = 2 
R2 24 
16 (2.4 V )
R3Vs
= 1.6 V
V3 =

24 
R3  R4
6.
a.
E 24 V
E
24 V
24 
= 12 A, I R2 =
=3A
=
=
=
R1 2 
R2  R3 6  + 2  8 
KCL: I + Is  I1  I R2 = 0
I1 =
I s = I1 + I R2  I = 12 A + 3 A  4 A = 11 A
b.
Vs = E = 24 V
VDR: V3 =
7.
68
R3 E
2 (24 V) 48 V
=6V
=
=
R2  R3 6  + 2  8 
a.
I=
E 22 V
= 4.68 A, Rp = Rs = 4.7 Ω
=
Rs 4.7 
b.
I=
E
9V
= 4.09 mA, Rp = Rs = 2.2 kΩ
=
Rs 2.2 k
CHAPTER 8
8.
9.
a.
E = IRs = (6 A)(12 Ω) = 72 V, Rs = 12 Ω
b.
E = IRs = (18 mA)(5.6 kΩ) = 100.8 V, Rs = 5.6 kΩ
a.
CDR: IL =
Es = IR = (20 A)(100 Ω) = 2 kV
Rs = 100 Ω
Es
2 kV
= 18.18 A
I=
=
Rs + RL 100  + 10 
b.
10.
11.
Rs ( I )
100 (20 A)
= 18.18 A, IL  I
=
Rs  RL 100  + 10 
a.
E = IR2 = (2 A)(5.6 Ω) = 11.2 V, R = 5.6 Ω
b.
ET = 12 V + 11.2 V = 23.2 V, RT = 10 Ω + 5.6 Ω = 15.6 Ω
c.
I3 =
a.
IT = 6.2 A  1.2 A  0.8 A = 4.2 A
b.
Vs = IT R = (4.2 A)(4 ) = 16.8 V
ET
23.2 V
= 217.64 mA

RT  91  15.6   91 
12.
IT = 7 A  3 A = 4 A
R ( I ) 6 (4 A)
= 2.4 A
CDR: I1 = 2 T =
R1  R2 4  + 6 
V2 = I1R1 = (2.4 A)(4 ) = 9.6 V
13.
a.
b.
Conversions: I1 = E1/R1 = 9 V/3 Ω = 3 A, R1 = 3 Ω
I2 = E2/R2 = 20 V/2 Ω = 10 A, R2 = 2 Ω
IT = 10 A  3A = 7 A, RT = 3 Ω  6 Ω  2 Ω
= 2 Ω  2 Ω
=1Ω
V ab = IT RT = (7 A)(1 Ω) = 7 V

14.
7V
= 1.17 A
6
c.
I3 =
a.
I=
b.
IT = 8 mA + 5.45 mA  3 mA = 10.45 mA
R = 6.8 k  2.2 k = 1.66 k
V1 = ITR = (10.45 mA)(1.66 k) = 17.35 V
CHAPTER 8
E
12 V
= 5.45 mA, Rp = 2.2 kΩ
=
R2 2.2 k 
69
15.
c.
V1 = V2 + 12 V  V2 = V1  12 V = 17.35 V  12 V
= 5.35 V
d.
I2 =
V2 5.35 V
= 2.43 mA
=
R2 2.2 k 
4  4I1  8I3 = 0
6  2I2  8I3 = 0
I1 + I2 = I3
────────────
a.
1
7
5
4
A, I 3  A
7
7
1
5
4
I R1  I1   A, I R2  I 2  A, I R3  I3  A
7
7
7
I1 =  A , I 2 
b.
16.
4 
Va = I3R3 =  A  (8 ) = 4.57 V
7 
10 + 12  3I3  4I1 = 0
12  3I3  12I2 = 0
I1 + I2 = I3
───────────────
a.
I1 = 3.06 A
I2 = 0.19 A
I3 = 3.25 A
I R1  I1 = 3.06 A, I R3  I 2 = 0.19 A = I12Ω
I R2  I 3 = 3.25 A
b.
I12Ω =
c.
17.
1.714 )(1.5 A)
= 0.19 A
1.714   12 
the same
10  I1 5.6 kΩ  I3 2.2 kΩ + 20 = 0
20 + I3 2.2 kΩ + I2 3.3 kΩ  30 = 0
I1 + I2 = I3
─────────────────────────
I1 = I R1 = 1.45 mA, I2 = I R2 = 8.51 mA, I3 = I R3 = 9.96 mA
70
CHAPTER 8
18.
1.2 kΩ I1 + 9  8.2 kΩ I3 = 0
10.2 kΩ I2 + 8.2 kΩ I3 + 6 = 0
I2 + I3 = I1
──────────────────────
a.
I1 = 2.03 mA, I2 = 1.23 mA, I3 = 0.8 mA
I R1 = I1 = 2.03 mA
I R2 = I3 = 0.8 mA
I R3  I R4 = I2 = 1.23 mA = I9.1kΩ
b.
V4 = I2R4 = (1.23 mA)(1.1 kΩ) = 1.35 V
Va = 6 V  V4 = 6 V  1.35 V = 4.65 V
19.
I1 = I R1 (CW), I2 = I R2 (down), I3 = I R3 (CW), I4 = I R4 (down)
I5 = I R5 (CW)
a.
E1  I1R1  I2R2 = 0
I2R2  I3R3  I4R4 = 0
I4R4  I5R5  E2 = 0
I1 = I2 + I3
I3 = I4 + I5
────────────────
c.
I2(R1 + R2) + I3R1
+ 0
= E1
 I3(R3 + R4) + I5R4
=0
I2(R2)
 I5(R4 + R5) = E2
0
+ I3R4
───────────────────────────
3I2 + 2I3 + 0 = 10
1I2  9I3 + 5I5 = 0
0 + 5I3  8I5 = 6
─────────────
d.
I3 = I R 3 = 63.69 mA (CW)
CHAPTER 8
b.
E1  I2(R1 + R2)  I3R1 = 0
I2R2  I3(R3 + R4) + I5R4 = 0
I3R4  I5(R4 + R5)  E2 = 0
───────────────────
71
20.
a.
b.
21
a.
4  4I1  8(I1  I2) = 0
8(I2  I1)  2I2  6 = 0
───────────────
5
1
I1 =  A , I2 =   A
7
7
1
I R1 = I1 =  A
7
5
I R2 = I2 =  A
7
4
 1   5 
I R3 = I1  I2 =   A     A  =
A (dir. of I1 )
7
 7   7 
4 
Va = I R3 R3   A  (8 ) = 4.57 V
7 
10  4I1  3(I1  I2)  12 = 0
12  3(I2  I1)  12I2 = 0
─────────────────
I1 = 3.06 A, I2 = 0.19 A
I E1 = 3.06 A (CCW)
I E2 = 3.06 A + 0.19 A = 3.25 A (up)
I R2 = I1  I2 = (3.06 A)  (0.19 A) = 3.25 A
b.
PE2  I E2 E2 = (3.25 A)(12 V) = 39 W
PR3  I R23 R3 = (0.19 A)2 12 Ω = 433.2 mW
22.
a.
10  I1(5.6 kΩ)  2.2 kΩ(I1  I2) + 20 = 0
20  2.2 kΩ(I2  I1)  I2 3.3 kΩ  30 = 0
────────────────────────────
I1 = 1.45 mA, I2 = 8.51 mA
I R1 = I1 = 1.45 mA, I R2 = I2 = 8.51 mA
I R3 = I2  I1 = 7.06 mA (direction of I2)
b.
23.
a.
V3.3kΩ = I2R2 = (8.51 mA)(3.3 kΩ) = 28.1 V
I1(1.2 kΩ) + 9  8.2 kΩ(I1  I2) = 0
I2(1.1 kΩ) + 6  I2 (9.1 kΩ)  8.2 kΩ(I2  I1) = 0
──────────────────────────────────
I1 = 2.03 mA, I2 = 1.23 mA
I R1 = I1 = 2.03 mA, I R3  I R4  I 2 = 1.23 mA
I R2  I1  I 2 = 2.03 mA  1.23 mA = 0.80 mA (direction of I1)
b.
72
Va = 6 V  I2(1.1 kΩ) = 6 V  (1.23 mA)(1.1 kΩ) = 6 V  1.35 V = 4.65 V
CHAPTER 8
24.
10  I12  1(I1  I2) = 0
1(I2  I1)  I2 4  5(I2  I3) = 0
5(I3  I2)  I3 3  6 = 0
──────────────────────
3I1  1I 2  0  10
a.
1I1  10 I 2  5 I 3  0
0  5I 2  8I 3   6
I1 = 3.31 A, I2 = 63.69 mA, I3 = 789.8 mA
b, c. Ignore
d.
I10V  = I1 = 3.31 A
I 6V  = I3 = (789.8 mA) = 789.8 mA
25.
a.
with
or
I1 2.2 kΩ  (I1  I2)9.1 kΩ + 18 V = 0
18 V  (I2  I1)9.1 kΩ  7.5 kΩ I2  (I2  I3)6.8 kΩ = 0
6.8 kΩ(I3  I2)  3 V  3.3 kΩ I3 = 0
──────────────────────
11.3 kΩ I1  9.1 kΩI2 = 18 V
23.4 kΩ I2  9.1 kΩI1  6.8 kΩ I3 = 18 V
10.1 kΩ I3  6.8 kΩ I2 = 3 V
──────────────────────
11.3 kΩ I1  9.1 kΩ I2
= 18 V
9.1 kΩ I2 + 23.4 kΩ I2  6.8 kΩ I3 = 18 V
6.8 kΩ I2 + 10.1 kΩ I3 = 3 V
─────────────────────────────
b.
I1 = 1.21 mA, I2 = 0.48 mA, I3 = 0.62 mA
c.
I E1  = I1  I2 = 1.21 mA  (0.48 mA) = 1.69 mA
I E2  = I3 = (0.62 mA) = 0.62 mA
26.
a.
16  4I1  3(I1  I2)  12  4(I1  I3) = 0
12  3(I2  I1)  10 I2  15  4(I2  I3) = 0
16  4(I3  I1)  4(I3  I2)  7I3 = 0
─────────────────────────────
b.
I1 = 0.24 A, I2 = 0.52 A, I3 = 1.28 A
c.
I R5 = I1 = 0.24 A
CHAPTER 8
73
27.
28.
d.
a.
6.8 k I1  4.7 k(I1  I2) + 6  2.2 k(I1  I4) = 0
6  4.7 k(I2  I1)  2.7 k I2  8.2 k (I2  I3) = 0
1.1 k I3  22 k(I3  I4)  8.2 k(I3  I2)  9 = 0
5  1.2 k I4  2.2 k(I4  I1)  22 k(I4  I3) = 0
────────────────────────────────────
b.
I1 = 0.03 mA, I2 = 0.88 mA, I3 = 0.97 mA, I4 = 0.64 mA
c.
I6V = I1  I2 = 0.03 mA  (0.88 mA) = 0.91 mA,
P6V = E I6V = (6 V)(0.91 mA) = 5.46 mW
a.
Network redrawn:
b.
2I1  6  4I1 + 4I2 = 0
4I2 + 4I1  1I2 + 1I3  6 = 0
1I3 + 1I2 + 6  8I3 = 0
c.
I1 = 3.8 A, I2 = 4.20 A, I3 = 0.20 A
PE2  E2 I 3 = (6 V)(0.2 A) = 1.2 W
PE1  E1 I 2 = (6 V)(4.2 A) = 25.2 W
PT  PE1  PE2 = 1.2 W + 25.2 W = 26.4 W
29.
74
a.
20 V  IB(270 kΩ)  0.7 V  IE(0.51 kΩ) = 0
IE(0.51 kΩ) + 8 V + IC(2.2 kΩ)  20 V = 0
IE = IB + IC
──────────────────────────────
IB = 63.02 μA, IC = 4.42 mA, IE = 4.48 mA
b.
VB = 20 V  IB(270 k) = 20 V  (63.02 A)(270 k) = 20 V  17.02 V = 2.98 V
VE = IERE = (4.48 mA)(510 ) = 2.28 V
VC = 20 V  IC(2.2 k) = 20 V  (4.42 mA)(2.2 k) = 20 V 9.72 V = 10.28 V
c.
  IC/IB = 4.42 mA/63.02 A = 70.14
CHAPTER 8
30.
24 V  6I1  4I2  10I1 + 12 V = 0
and 16I1 + 4I2 = 36
I1  I2 = 6 A
───────────────────
I1 = I2 + 6 A
16[I2 + 6 A] + 4I2 = 36
16I2 + 96 + 4I2 = 36
20I2 = 60
I2 = 3 A
I1 = I2 + 6 A = 3 A + 6 A = 3 A
I24V = I6 = I10 = I12V = 3 A (CW)
I4 = 3 A (CCW)
31.
20 V  4I1  6(I1  I2)  8(I3  I2)  1I3 = 0
10I1  14I2 + 9I3 = 20
I3  I1 = 3 A
I2 = 8 A
────────────────────────
10I1  14(8 A) + 9[I1 + 3 A] = 20
19I1 = 105
I1 = 5.526 A
I3 = I1 + 3 A = 5.526 A + 3 A = 8.526 A
I2 = 8 A
I20V = I4Ω = 5.53 A (dir. of I1)
I6 = I2  I1 = 2.47 A (dir. of I2)
I8 = I3  I2 = 0.53 A (dir. of I3)
I1 = 8.53 A (dir. of I3)
CHAPTER 8
75
32.
a.
b.
33.
a.
b.
34.
(4 + 8)I1  8I2 = 4
(8 + 2)I2  8I1 = 6
─────────────
 1   5  4
I8  I1  I 2    A     A   A
 7   7  7
(4 + 3)I1  3I2 = 10  12
(3 + 12)I2  3I1 = 12
─────────────────
I 3   I 2  I1  0.19 A  (3.06 A) = 3.25 A
a.
a.
I1(5.6 k + 2.2 k)  2.2 k (I2) = 10 + 20
I2(2.2 k + 3.3 k)  2.2 k (I1) = 20  30
──────────────────────────────
b.
I E1 = I1 = 1.45 mA, I E2  = 8.51 mA,
I E3  = I1  I2 = (1.45 mA)  (8.5 mA) = 9.96 mA
35.
36.
a.
b.
I1(2 + 1)  1I2 = 10
I2(1 + 4 + 5)  1I1  5I3 = 0
I3(5 + 3)  5I2 = 6
───────────────────
I1 = 3.31 A, I2 = 63.69 mA, I3 = 789.8 mA
c.
I R2  I1  I 2 = (3.31 A)  (63.69 mA) = 3.37 A
a.
b.
(2.2 k + 9.1 k)I1  9.1 kI2 = 18
(9.1 k + 7.5 k + 6.8 k)I2  9.1 k I1  6.8 kI3 = 18
(6.8 k + 3.3 k)I3  6.8 kI2 = 3
───────────────────────────
I1 = 1.21 mA, I2 = 0.48 mA, I3 = 0.62 mA
c.
I E1  = I1  I2 = 1.21 mA  (0.48 mA) = 1.69 mA
I E2  = I3 = (0.62 mA) = 0.62 mA
76
CHAPTER 8
37.
(3 Ω + 6 Ω)I1  6 Ω I2 = 9 V
(6 Ω + 2 Ω)I2  6 Ω I1 = 20 V
────────────────────
a.
9 Ω I1  6 Ω I2 = 9 V
6 Ω I1 + 8 Ω I2 = 20 V
────────────────────
I1 = 5.33 A, I2 = 6.5 A
b.
38.
39.
Vab = 2 Ω (I2)  20 V = 2 Ω = 2 Ω(6.5 A)  20 V = 13 V  20 V = 7 V
a.
I1(6.8 k + 4.7 k + 2.2 k)  4.7 k I2  2.2 k I4 = 6
I2(2.7 k + 8.2 k + 4.7 k)  4.7 k I1  8.2 k I3 = 6
I3(8.2 k + 1.1 k + 22 k)  22 k I4  8.2 k I2 = 9
I4(2.2 k + 22 k + 1.2 k)  2.2 k I1  22 k I3 = 5
───────────────────────────
b.
I1 = 0.03 mA, I2 = 0.88 mA, I3 = 0.97 mA, I4 = 0.64 mA
c.
I 22k  = I4  I3 = (0.64 mA)  (0.97 mA) = 0.33 mA
V22kΩ = I22kΩ  22 kΩ = (0.33 mA)(22 kΩ) = 7.26 V
a.
(1 Ω + 2 Ω + 4 Ω)I1  2 ΩI2  4 ΩI3 = 12 V
(2 Ω + 2 Ω + 10 Ω)I2  2 ΩI1  10 ΩI3 = 20 V
(4 Ω + 10 Ω + 8 Ω)I3  10 ΩI2  4 ΩI1 = 20 V
───────────────────────────
7I1  2I2  4I3 = 12
2I1  14I2 + 10I3 = 20
4I1 + 10I2  22I3 = 20
────────────────
b.
I1 = 2.38 A, I2 = 0.195 A, I3 = 1.25 A
c.
Va = (I1  I3)4 Ω = (2.38 A  1.25 A)4 Ω = 4.5 V
Vb = I38 Ω = (1.25 A)(8 Ω) = 10 V
d.
Vab = Va  Vb = 4.5 V  10 V = 5.5 V
CHAPTER 8
77
40.
a.
At V1:  I i   I o
0
V1
V V
 5A  1 2
2
8
 Ii   Io
At V2:
V1  V2
V
3A 2
8
4
and
41.
1 1
1
V1     V2    5
 2 8
8
1
1 1 
V1    V2     3
8
8 4 
───────────────────────────
b.
V1 = 10.27 V, V2 = 11.36 V
c.
V8Ω = V1  V2 = 10.27 V  (11.36 V) = 1.09 V
d.
I 2  =
V1 10.27 V

= 5.14 A
2
2
V
11.36 V
I 4  = 2 
= 2.84 A
4
4
a.
At V1:  I i   I o
V1
 12 A  I 6  and V1  I 6  54 V  V2  0
8
V  V  54 V V1
V
or I = 1 2

 2 9 A
6
6 6
V1
V1
V2
 12 A 

 9A
so that 0 =
8
6 6
1 
 1
 1 
or V1 

 V2 
= 12 A + 9 A = 3 A

8  6  
 6  
0
At V2:  I i   I o
V2
V
 2
20  5 
V1
V2
V
V

9 A  2  2
6 6
20  5 
I
or
1
1 
 1
 1 
and V2 


 V1 
= 9 A

 6 
 6  20  5  
78
CHAPTER 8
42.
b.
1 
 1
 1 
resulting in V1 

 V2 
= 3 A

 6 
8  6 
1
1 
 1 
 1
V1 
 V2 


= 9 A

 6 
 6  20  5  
──────────────────────────────
V1 = 29.29 V, V2 = 33.34 V
c.
I 20   
V2
33.34 V

= 1.67 A
20 
20 
a.
At V1:  I i   I o
V1 V1  V2

2A
2
4
4A
Ar V2:  I i   I o
2A
or
43.
V1  V2
V
V
 2  2
4
20  5 
1 1
1
V1     V2    2
2 4
 4
1
1 1 1
V1    V2  
 2
4
 4 20 5 
b.
V1 = 4.8 V, V2 = 6.4 V
c.
I1: P = V1I1 = (4.8 V)(4 A) = 19.2 W
I2: P = (V1  V2 ) I 2  (4.8 V  6.4 V)(2 A) = 3.2 W
a.
At V1:  I i   I o
0  6A 
V1 V1  V2 V1  V2


5
3
2
At V2:  I i   I o
7A 
V1  V2 V1  V2 V2
V


 2
3
2
4 8
1
1 
1 
 1
 1
so that V1 
= 6 A


 V2 


 3  2  
5  3  2 
1
1
1 
1 
 1
 1
V2 
=7A



 V1 


 3  2  
4  8  3  2 
────────────────────────────────────
or 1.03V1  0.833V2 = 6
0.833V1 + 1.21V2 = 7
───────────────────
CHAPTER 8
79
44.
b.
V1 = 2.59 V, V2 = 4 V
c.
V2Ω = V3Ω = V2  V1 = 4 V  (2.59 V) = 6.59 V
V5Ω = V1 = 2.59 V
V4Ω = V8Ω = V2 = 4 V
a.
Source conversion: I3 =
12 V
= 3 A, Rp = R3 = 4 Ω
4
At V1:  I i   I o
0
V1
V
V V
 1 5 A 1 2 + 3 A
3 6
4
At V2:  I i   I o
3A 
V1  V2 V2

4A
4
8
Rewritten:
b.
c.
45.
1
1  V2
 1
V1 



= 5 A  3 A
 3  6  4   4 
1 
 1 
 1
V1 
 V2 

= 4 A + 3A

 4  8 
4 
V1 = 14.86 V, V2 = 12.57 V
14.86 V
I 6  
= 2.48 A
6
a.
Source Conversion: I2 =
At V1:  I i   I o
5A
15 V
= 5 A, Rp = R1 = 3 Ω
3
V1 V1  V2 V1  V3


3
6
6
At V2:  I i   I o
V  V3
V1  V2 V2

3 A 2
6
4
5
At V3:  I i   I o
V1  V3 V2  V3 V3


6
5
7
80
CHAPTER 8
Rewritten:
1
1 
1
1
 1
V1 
V2 
V3 = 5 A









3
6
6
6
6


1
1 
1
1
 1
V2 



V1 
V3 = 3 A






6
4
5
6
5


1
1 
1
1
 1
V3 
V2 
V1 = 0









6
5
7
5
6


────────────────────────────────────
46.
b.
V1 = 7.24 V, V2 = 2.45 V, V3 = 1.41 V
c.
V5 = V3  V2 = 1.41 V  (2.45 V) = 3.86 V
 +
a.
Source Conversion: I2 =
16 V
= 4 A, Rp = R2 = 4 Ω
4
At V1:  I i   I o
02 A+
V1 V1  V3 V1  V2


9  20 
20 
At V2:  I i   I o
V1  V2 V2  V3
V

 2
20 
20  18 
At V3:  I i   I o
V1  V3 V2  V3
V

 4A  3
20 
20 
4
Rewritten:
1
1 
1
1
 1
V1 



V2 
V3 = 2 A

20 
 9  20  20   20 
1
1 
1
1
 1
V2 



V1 
V3 = 0

20 
 20  20  18   20 
1
1 
1
1
 1
V3 
V2 
V1 = 4A




20 
 20  20  4   20 
────────────────────────────────────
CHAPTER 8
b.
V1 = 6.64 V, V2 = 1.29 V, V3 = 10.66 V
c.
VR6 = V3  V1 = 10.66 V  (6.64 V) = 17.30 V
 +
81
47.
a.
At V1:  I i   I o
05 A
V1 V1  V2

2
2
At V2:  I i   I o
V  V3
V1  V2 V2
V

 2  2
2
9 7
2
At V3:  I i   I o
V2  V3
V
V
5 A  3  3
2
2 4
Rewritten:
1 
1
 1
 2   2   V1  2  V2  0 = 5 A
1
1
1 
1
1
 1
 2   9   7   2   V2  2  V1  2  V3 = 0
1
1 
1
 1
V3 


V3 
V2 = 5A

2
2  2  4 
────────────────────────────────────
48.
b.
V1 = 5.31 V, V2 = 0.62 V, V3 = 3.75 V
c.
I9 
V2
0.62 V

= 68.9 mA
9
9
a.
At V1:  I i   I o
05 A+
V1 V1  V3

2
6
At V2:  I i   I o
5A2 A
V2
4
At V3:  I i   I o
V1  V3
V
 2A  3
6
5
82
CHAPTER 8
Rewritten:
1 
1
 1
V1 


V3 = 5 A




2
6
6


 1 
V2 
=5A2A
 4  
1 
1
 1
V3 
V1 = 2A






6
5
6


───────────────────────
49.
b.
V1 = 6.92 V, V2 = 12 V, V3 = 2.3 V
c.
I4Ω =
V2 12 V

=3A
4 4
a.
 Ii   Io
Node V1:
2A=
V1 V1  V2

6  10 
Supernode V2, V3:
V
V V
V
0= 2 1  2  3
10 
4  12 
Independent source:
V2  V3 = 24 V or V3 = V2  24 V
2 eq. 2 unknowns:
V1 V1  V2

=2A
6  10 
V2  V1 V2 V2  24 V


=0
10 
4
12 
─────────────────────
0.267V1  0.1V2 = 2
+0.1V1  0.433V2 = 2
────────────────
V1 = 10.08 V, V2 = 6.94 V
V3 = V2  24 V = 17.06 V
CHAPTER 8
83
50.
 Ii   Io
Supernode:
3A+4A=3A+
V1
V
 2
20  40 
V1
V

 2
4 A 
2 eq. 2 unk. 
20  40 
V2  V1  16 V

Subt. V2 = 16 V + V1
V
(16 V  V1 )
4A= 1 
20 
40 
and V1 = 48 V
V2 = 16 V + V1 = 64 V
──────────────────
51.
a.
1 1
1
V1     V2   = 5
 2 8
8
1
1 1 
V1    V2    = 3
8
8 4 
────────────────────
V1 = 10.27 V, V2 = 11.36 V
b.
52.
VI1 = V1 = 10.27 V, VI 2 = V2 = 11.36 V
a.
1 1 
1
V1     V2   = 12 A + 9 A = 3 A
8
6


6
 1 1 1
1
V2      V1   = 9 A
20
5
6


6
──────────────────────────────
V1 = 29.29 V, V2 = 33.34 V
84
CHAPTER 8
b.
53.
V1  V6Ω  54 V  V2 = 0
V6 = V1  V2  54 V = 29.29 V  (33.34 V)  54 V = 49.95 V
+ 
a.
1
1 
1
1
 1
V1 



V2 
V3 = 2 A

20 
 9  20  20   20 
1
1 
1
1
 1
V2 



V1 
V3 = 0

20 
 20  20  18   20 
1
1 
1
1
 1
V3 
V2 
V1 = 4A




20 
 20  20  4   20 
V1 = 6.64 V, V2 = 1.29 V, V3 = 10.66 V
b.
54.
Original Network:
V4Ω = 16 V  V3 = 16 V  10.66 V = 5.34 V
5.34 V
I 4Ω =
= 1.34 A
4
a.
1 
1
 1
 2   2   V1  2  V2  0 = 5 A
1
1
1 
1
1
 1
 2   9   7   2   V2  2  V1  2  V3 = 0
1
1 
1
 1
 2   2   4   V3  2  V2 = 5A
────────────────────────────────────
CHAPTER 8
b.
V1 = 5.31 V, V2 = 0.62 V, V3 = 3.75 V
c.
I9 
V2
0.62 V

= 68.89 mA
9
9
85
55.
56.
1 
1
 1
V1 


V3 = 5 A

 2  6  6 
 1 
V2 
=5A2A
 4  
1 
1
 1
V3 
V1 = 2A



6  5  6 
───────────────────────
a.
b.
V1 = 6.92 V, V2 = 12 V, V3 = 2.3 V
c.
I2Ω =
V2
6.92 V

= 3.46 A
2
2
a.
1
1 
1
1
 1
V1 



V2 
V3 = 12 A

2
1  2  2   2 
1
1 
1
1
 1
V2 



V1 
V3 = 2 A

10 
 2  4  10   2 
1
1 
1
1
 1
V3 



V1 
V2 = 2A

10 
 2  10  8   2 
────────────────────────────────────
and
2V1  0.5V2  0.5V3 = 12
0.5V1 + 0.85V2  0.1V3 = 2
0.5V1  0.1V2 + 0.725V3 = 2
───────────────────────
V1 = 9.63 V, V2 = 4.49 V, V3 = 10.02 V
b.
57.
Vab = V2  V3 = 4.49 V  10.02 V = 5.53 V
a.
I1(6 Ω + 2 Ω + 10 Ω)  2 ΩI2  10 ΩI3 = 12 V
I2(2 Ω + 5 Ω + 5 Ω)  2 ΩI1  5 ΩI3 = 0
I3(5 Ω + 20 Ω + 10 Ω)  10 ΩI1  5 ΩI2 = 0
or
18I1  2I2  10I3 = 12
2I1 + 12I2  5I3 = 0
10I1  5I2 + 35I3 = 0
───────────────
I1 = 850.99 mA, I2 = 258.53 mA, I3 = 280.07 mA
86
CHAPTER 8
58.
b.
I R5 = I3  I2 = 280.07 mA  258.53 mA = 21.54 mA
c.
no
d.
no 2 Ω/10 Ω =
1
1
 5 Ω/20 Ω =
5
4
a.
1
1 
1
1
 1
V1 



V2 
V3 = 2 A

5
6  2  5  2 
1
1 
1
1
 1
V2 
V1 
V3 = 0




5
 2  5  10   2 
1
1 
1
1
 1
V3 
V2 
V1 = 0




5
 5  5  20   5 
────────────────────────────────────
or 0.867V1  0.5V2  0.2V3 = 12
0.5V1 + 0.8V2  0.2V3 = 0
0.2V1  0.2V2 + 0.45V3 = 0
───────────────────────
V2 = 5.7 V, V3 = 5.6 V
59.
b.
V5Ω = V2  V3 = 5.7 V  5.6 V = 0.1 V
c.
no
d.
no 2 Ω/10 Ω =
1
1
 5 Ω/20 Ω =
5
4
a.
Source conversion: E = IR = (12 m)(2 kΩ) = 24 V
Rs = 2 kΩ
I1(2 kΩ + 33 kΩ + 3.3 kΩ)  33 kΩI2  3.3 kΩI3 = 24 V
I2(33 kΩ + 56 kΩ + 36 kΩ)  33 kΩI1  36 kΩI3 = 0
I3(36 kΩ + 3.3 kΩ + 5.6 kΩ)  3.3 kΩI1  36 kΩI2 = 0
────────────────────────────────────
I1 = 0.97 mA, I2 = I3 = 0.36 mA
b.
I5 = I2  I3 = 0.36 mA  0.36 mA = 0 mA
c, d. yes
CHAPTER 8
87
60.
a.
1
1
1
1 
 1
V1 

V2 
V3 = 12 mA



56 k
 2 k 33 k 56 k 33 k
1
1
1
1 
 1
V2 

V1 
V3 = 0



36 k
 33 k 3.3 k 36 k 33 k
1
1
1
1 
 1

V3 
V2 
V1 = 0



56 k
 56 k 36 k 5.6 k  36 k
────────────────────────────────────
Rewritten:
548.16V1  30.3V2  17.86V3 = 12 × 103
30.3V1 + 361.11V1  27.78V3 = 0
17.86V1  27.78V2 + 224.21V3 = 0
────────────────────────────
V2 = 2.01 V, V3 = 2.01 V
b.
VR5 = V2  V3 = 2.01 V  2.01 V = 0 V
c, d. yes
61.
Mesh Analysis
(1 k + 2 k + 2 k)I1  2 k I2  2 k I3 = 10
(2 k + 2 k + 2 k)I2  2 k I1  2 k I3 = 0
(2 k + 2 k + 2 k)I3  2 k I1  2 k I2 = 0
─────────────────────────────────
I1 = I10V = 3.33 mA
Nodal Analysis:
Source conversion: I = 10 V/1 kΩ = 10 mA, R = 1 kΩ
1
1 
1
1
 1

V1 
V2 
V3 = 10 mA
+
+

2 k
 1 k 2 k 2 k  2 k
1
1 
1
1
 1
V2 
V1 
V3 = 0
+
+


2 k
 2 k 2 k 2 k  2 k
1
1 
1
1
 1
V3 
V2 
V1 = 0
+
+


2
k
2
k
2
k
2
k
2
k




 
──────────────────────────────────────
V1 = 6.67 V = E  IRs = 10 V  I(1 k)
10  6.67 V
= 3.33 mA
I=
1k
88
CHAPTER 8
62.
Mesh Analysis
Source conversion: E = 20 V, R = 10 
(10 + 10 + 20)I1  10I2  20I3 = 20
(10 + 20 + 20)I2  10I1  20I3 = 0
(20 + 20 + 10)I3  20I1  20I2 = 0
──────────────────────
I1 = I20V = 0.83 A
V  20 V  8.3 V
= 11.7 V
Is =
V 11.70 V
= 1.17A

10 
Rs
Nodal Analysis:
1
1 1
1
1
V1        V2    V3  2
 10 10 20   20 
10 
1
1 1
1
1
V2  
     V1    V3  0
 20 20 10   20 
 20 
1
1 1
1
1
V3  
     V1    V2  0
10 20 20  10 
 20 
───────────────────────
V
I Rs = 1 = 1.17 A
Rs
63.
I=
=
20 V
4
2
 2

 +   + 3     + 4 
5
5
 5

20 V
4
 + (3.14 )  (4.4 )
5
= 7.36 A
CHAPTER 8
89
64.
RT = 2.27 k + [4.7 k + 2.27 k]  [1.1 k + 2.27 k]
= 2.27 k + [6.97 k]  [3.37 k]
= 2.27 k + 2.27 k
= 4.54 k
8V
= 1.76 mA
I=
4.54 k
(Y-Δ conversion)
400 V
400 V

I=
12 k  12 k  6 k
3 k
= 133.33 mA
65.
66.
a.
I=
42 V
42 V

(18   18 )   (18   18 )  (18   18 ) 9   9   9  
= 7 A (YΔ conversion)
b.
  Y conversion
I s1 =
10 V
5V
15 V
+
=
= 0.83 mA
18 k  18 k  18 k 
67.
90
CHAPTER 8
68.
a.
b.
69.
R = R1 + 1 k = 3 k
R = R2 + 1 k = 3 k
3k
= 1.5 k
RT =
2
RT = 1 k + 1.5 k + 1 k = 3.5 k
E
20 V
= 5.71 mA
Is =
=
RT 3.5 k 
Using two   Y conversions:
c  g: 27   9   27  = 5.4 
a  h: 27   9   27  = 5.4 
RT = 5.4   (13.5  + 5.4 )
= 5.4   18.9
= 4.2 
CHAPTER 8
91
Chapter 9
1.
a.
RT (from source) = 4 Ω + 2 Ω  12 Ω
= 4 Ω + 1.71 Ω
= 5.71 Ω
E1
16 V
Is 

 2.8 A
RT 5.71 V
2 (2.8 A)
 0.4 A
I12  
2   12 
E1:
RT (from source) = 2 Ω + 4 Ω  12 Ω
=2Ω+3Ω
=5Ω
E2 10 V
Is 

2A
RT
5
4 (2 A)
 0.5 A
I12  
4   12 
E2:
I12Ω = 0.5 A  0.4 A = 0.1 A
b.
I12 
c.
2.
1.333 (1A)
= 0.1 A
1.333   12 
the same
a.
24 (3 A)
= 2.25 A
24   8 
V   I R = (2.25)(4.7 Ω) = 10.575 V
I 
V  
4.7 (12 V)
= 1.763 V
4.7   3.3   24 
V   10.575 V  1.763 V = 8.81 V

92
CHAPTER 9
b.
c.
3.
V  2 (10.575 V)2
= 23.79 W

R
4.7 
V  2 (1.763 V) 2
P=
= 0.661 W

R
4.7 
P=
V 2 (8.81 V) 2

= 16.51 W
R
4.7 
d.
P=
e.
23.79 W + 0.661 W  16.51 W
24.45 W  16.51 W
E:
RT = 12   24   56   28.8 
E
24 V
Is =
= 0.833 A

RT 28.8 
24 (0.833 A)

I 56
= 0.25 A
24   56 
I:
24   56   16.8 
12 (8 A)
I  
= 3.33 A
12   16.8 
24 (3.33 A)
  
I 56
=1A
24   56 
I 56   I   I   0.25 A+1 A = 1.25 A
4.
E1:
42 V
= 1.944 A
18  + 3.6 
9 ( IT ) 9 (1.944 A)
I1 =

96
15 
= 1.17 A
IT =
E2:
IT =
E2 24 V
=2A

RT 12 
I24V = IT + I1 = 2 A + 1.17 A = 3.17 A (dir. of I1)
CHAPTER 9
93
5.
E:
V2 =
6.8 k(36 V)
= 13.02 V
6.8 k  12 k
I:
I2 =
12 k(9 mA)
= 5.75 mA
12 k  6.8 k
V2  I 2 R2 = (5.75 mA)(6.8 k) = 39.10 V
V2 = V2  V2 = 13.02 V + 39.10 V = 52.12 V
6.
1.2 k  4.7 k  0.956 k
3.3 kΩ + 0.956 kΩ = 4.256 kΩ
4.256 k(5 mA)
I 
2.2 k  4.256 k
 3.3mA
I:
E:
I  I   I  = 3.3 mA + 0.986 mA = 4.286 mA
7.
E1:
I1 =
94
2.2 kΩ + 3.3 kΩ = 5.5 k Ω
5.5 k  4.7 k  2.53 k
RT = 2.54 kΩ + 1.2 kΩ = 3.73 kΩ
8V
Is 
= 2.14 mA
3.73 k
4.7 k(21.4 mA)
I  
= 0.986 mA
4.7 k  5.5 k
E1
12 V
= 1.03 A

RT 6   5.88 
CHAPTER 9
30 ( I1 )
30 (1.03 A)

30   7 
37 
 835.14 mA
Vs = I(4 ) = (835.14 mA)(4 )
= 3.34 V
I 
I:
8 (6 A)
=4A
8  4
Vs = I(4 ) = 4 A(4 ) = 16 V
I =
E2:
RT = 12   (4  + 5 ) = 12   9  = 5.14 
E
8V
I = 2 
 0.875 A
RT 4   5.14 
12 ( I )
12 (0.875 A)
I =

 0.5 A
12   9 
21 
Vs I (4 ) = 0.5 A(4 ) = 2 V
Vs = Vs Vs  Vs = 16 V  3.34 V  2 V = 10.66 V
8.
a.
RTh = R3 + R1  R2 = 4 Ω + 6 Ω  3 Ω = 4 Ω + 2 Ω = 6 Ω
R2 E
3 (18 V)
ETh =

6V
R2  R1 3   6 
CHAPTER 9
95
9.
b.
I1 =
a.
RTh:
ETh
6V
= 0.75 A
=
RTh  R 6  + 2 
6V
= 166.67 mA
I2 =
6  + 30 
6V
I3 =
= 56.60 mA
6  + 100 
 RTh  3.3 k  1.2 k  2.4 k
 3.3 k  0.8 k
 4.1 k
ETh:
ETh  (120 mA)(2.4 k  1.2 k)
 96 V
RTh = 4.1 kΩ
b.
96 V
= 15.74 mA
6.1 k
P = I2R = (15.74 mA)2 2 kΩ = 0.495 W
R = 100 kΩ:
96 V
= 0.922 mA
I=
104.1 k
P = I2R = (0.922 mA)2 100 kΩ = 85 mW
I=
10.
a.
RTh:
 RTh = 5 Ω + 5 Ω  5 Ω = 7.5 Ω
ETh:
ETh =
96
20 V
= 10 V
2
CHAPTER 9
2
b.
 ETh 
 10 V

R = 2 Ω: P = 
R =
 2  = 2.22 W
R

R
7.5


2



 Th

2
2
10 V


R = 100 Ω: P = 
 100  = 0.87 W
7.5


100



11.
RTh = 3 Ω  8 Ω = 2.18 Ω
18 V + 12 V
= 2.73 A
3 8 
V3Ω = IR = (2.73 A)(3 Ω) = 8.19 V
ETh = 18 V  8.19 V = 9.81 V
I=
12.
RTh:
RTh = 5.6 kΩ  2.2 kΩ = 1.58 kΩ
ETh: Superposition:
I:
ETh = IRT
= 8 mA(5.6 k  2.2 k)
= 8 mA(1.579 k)
= 12.64 V
E:
5.6 k(16 V)
5.6 k  2.2 k
= 11.49 V
ETh =
+
ETh = 11.49 V  12.64 V = 1.15 V

CHAPTER 9
97
13.
RTh:
RTh = 4   (2  + 6 Ω  3 ) = 2 
ETh:
72 V
=9A
6   3   (2   4 )
3 ( IT ) 3 (9 A)
=3A
I2 =

3  6
9
ETh = V6 + V2 = (IT)(6 ) + I2(2 )
= (9 A)(6 ) + (3 A)(2 ) = 60 V
IT =
14.
a.
RTh:
RTh = 2.7 k  (4.7 k + 3.9 k) = 2.7 k  8.6 k = 2.06 k
ETh:
I =
3.9 k(18 mA)
= 6.21 mA
3.9 k  7.4 k
ETh = I(2.7 k) = (6.21 mA)(2.7 k) = 16.77 V
b.
16.77 V
16.77 V

2.06 k  1.2 k 3.26 k
= 5.14 mA
I=
98
CHAPTER 9
15.
a.
RTh:
RTh = 2  + 8  = 10 
ETh:
ETh = V16
20 V
= 825.08 mA
20   4.24 
5 ( IT )
5 (825.08 mA)
= 125.01 mA
I =

5   28 
33 
ETh = V16Ω = (I)(16 Ω) = (125.01 mA)(16 Ω) = 2 V
IT =
b.
16.
a.
20 Ω:
ETh
2V
2V
= 66.67 mA


RTh  R 10   20  30 
2V
2V

= 33.33 mA
50 Ω: I =
10   50  60 
2V
2V
100 Ω: I =

= 18.18 mA
10   100  110 
I =
RTh:
RTh = 3. 3 k + 2.2 k  1.1 k
= 3.3 k + 0.73 k
= 4.03 k
ETh: Superposition:
E1:
ETh = V2.2kΩ =
2.2 k(12 V)
2.2 k  1.1 k
=8V
CHAPTER 9
99
ETh = E2 = 4 V
ETh = ETh + ETh = 8 V + 4 V = 12 V
b.
V=
17.
1.2 k(12 V)
= 2.75 V
1.2 k  4.03 k
RTh:
RTh = 2.2 k  5.6 k = 1.58 k
R = 1.58 k + 3.3 k
= 4.88 k
R = 4.88 k  6.8 k = 2.84 k
RTh = 1.2 k + R = 1.2 k + 2.84 k = 4.04 k
ETh: Source conversions:
22 V
= 10 mA, Rs = 2.2 k
I1 =
2.2 k
12 V
I2 =
= 2.14 mA, Rs = 5.6 k
5.6 k
Combining parallel current sources: IT = I1  I2 = 10 mA  2.14 mA = 7.86 mA
2.2 k  5.6 k = 1.58 k
100
CHAPTER 9
Source conversion:
E = (7.86 mA)(1.58 k) = 12.42 V
R = Rs + 3.3 k = 1.58 k + 3.3 k = 4.88 k
I=
12.42 V  6 V
6.42 V
= 549.66 A

4.88 k  6.8 k 11.68 k
V6.8kΩ = I(6.8 k) = (549.66 A)(6.8 k) = 3.74 V
ETh = 6 V + V6.8kΩ = 6 V + 3.74 V = 9.74 V
18.
a.
RTh:
RTh = 51 k  10 k = 8.36 k
ETh:
ETh =
b.
10 k(20 V)
= 3.28 V
10 k  51 k
IERE + VCE + ICRC = 20 V
but IC = IE
and IE(RC + RE) + VCE = 20 V
20 V  VCE
20 V  8 V
12 V
= 4.44 mA
or IE =


2.2 k  0.5 k 2.7 k
RC  RE
c.
ETh  IBRTh  VBE  VE = 0
E  VBE  VE 3.28 V  0.7 V  (4.44 mA)(0.5 k)
and IB = Th

8.36 k
RTh
2.58 V  2.22 V 0.36 V
=
= 43.06 μA
=
8.36 k
8.36 k
CHAPTER 9
101
19.
d.
VC = 20 V  ICRC = 20 V  (4.44 mA)(2.2 k)
= 20 V  9.77 V
= 10.23 V
a.
ETh = 20 V
I = 1.6 mA =
ETh 20 V
20 V
= 12.5 k
, RTh 

1.6 mA
RTh
RTh
b.
ETh= 60 mV, RTh = 2.72 k
c.
ETh = 16 V, RTh = 2.2 k
20.
RTh = 4   (2   2 ) 
4
=2
2
2 (6 V)
12 V

= 1.5 V
2 4 2  8 
V2Ω = V4Ω = 1.5 V
ETh = V4Ω + V2Ω = 1.5 V + 1.5 V = 3 V
V4Ω =
21.
a.
From Problem 8, RN = RTh = 6 Ω
RT  6   3   4 
 6   1.714   7.714 
E
18 V
Is =
= 2.333 A

RT 7.714 
3 (2.333 A)
=1A
IN =
3 4 
102
b.
RTh = 6 Ω, ETh = INRN = (1 A)(6 Ω) = 6 V
c.
same results
CHAPTER 9
22.
a.
From Problem 9, RN = RTh = 4.1 kΩ
2.4 k(120 mA)
2.4 k  (1.2 k  3.3 k)
 87.80 mA
I 
1.2 k(87.80 mA)
1.2 k  3.3 k
 23.41 mA
RTh = 4.1 kΩ, ETh = INRN = (23.41 mA)(4.1 kΩ) = 96 V
same results.
IN 
b.
c.
23.
From Problem 11, RN = RTh = 2.18 Ω
IN = 6 A  1.5 A = 4.5 A
24.
From Problem 12, RN = RTh = 1.58 kΩ
IN = 8 mA  7.27 mA = 0.73 mA
25.
From Problem 13, RN = RTh = 2 Ω
72 V
= 18 A
4
72 V
I3Ω =
= 16 A
3 6 2 
6 (16 A)
I2Ω =
= 12 A
6 2 
IN = I4Ω + I2Ω = 18 A + 12 A = 30 A
I4Ω =
CHAPTER 9
103
26.
From Problem 15, RN = RTh = 10 Ω
RT = 20 Ω + 5 Ω  (12 Ω + 1.778 Ω) = 23.67 Ω
E
20 V
Is =
= 844.95 mA

RT 23.67 
5 (844.95 mA)
=224. 98 mA
I12Ω =
5   (12   1.778 )
16 (224.98 mA)
IN =
= 200 mA
16   2 
27.
104
From Problem 17, RN = RTh = 4.04 kΩ
CHAPTER 9
4.88 k(3.427 mA)
4.88 k  1.02 k
 2.83 mA
I 
IN =
28.
6.8 k(2.83 mA)
= 2.41 mA
6.8 k   k
From Problem 20, RN = RTh = 2 Ω
IN 
29.
a.
6V
= 1.5 A
4
RN:
RN = 4   12  = 3 
E = 12 V:
IN =
CHAPTER 9
12 V
=3A
4
105
I = 2 A:
IN = 2 A
IN = IN + IN = 3 A + 2 A = 5 A
b.
I:
3 (5A)
3    
 145.63 mA
I 
V   IR
 (145.63 mA)(100 )
 14.56 V
E:
100 (72 V)
100   3 
 69.9 V
V100   V   V 
V  
 69.9 V  14.56 V
 55.34 V
30.
31.
106
a.
R = RTh = 6 Ω from Problem 8
b.
ETh = 6 V from Problem 8
E2
(6 V)2
= 1.5 W
Pmax = Th 
4 RTh 4(6 )
a.
R = RTh = 2.18 Ω from Problem 11
b.
ETh = 9.81 V from Problem 11
2
ETh
(9.81 V) 2
= 11.06 W
Pmax =

4 RTh 4(2.18 )
CHAPTER 9
32.
33.
34.
a.
R = RTh = 2 Ω from Problem 13
b.
ETh = 60 V from Problem 13
E2
(60 V) 2
= 450 W
Pmax = Th 
4 RTh 4(2 )
a.
R = RTh = 4.04 kΩ from Problem 17
b.
ETh = 9.74 V from Problem 17
E2
(9.74 V)2
= 5.87 mW
Pmax = Th 
4 RTh 4.04 k)
a.
R = RN = RTh = 2.18 Ω
b.
Pmax =
I N2 RN (13.33A)2 2.18 

= 96.84 W
4
4
2
35.
36.
 ETh 
Pmax = 
 R4
 RTh  R4 
with R1 = 0 Ω, ETh is a maximum and RTh is a minimum.
 R1 = 0 Ω
a.
V, and therefore V4 wll be its largest value
when R2 is as large as possible. Therefore,
choose R2 = open-circuit ( Ω) and
V2
P4 = 4 will be a maximum.
R4
b.
37.
No, examine each individually.
The voltage VL will be a maximum when R = 500 Ω because the full voltage, E, will appear
across RL.
Pmax =
CHAPTER 9
VL2 E 2 (12 V) 2
= 1.44 W


RL RL
500 
107
IT = 4 A + 7 A = 11 A
RT = 10   6   3  = 1.67 
VL = ITRT = (11 A)(1.67 ) = 18.37 V
V
18.37 V
= 6.12 A
IL = L 
RL
3
38.
39.
5 V / 2.2 k  20 V / 8.2 k
= 0.2879 V
1/ 2.2 k  1/ 8.2 k
1
Req =
= 1.7346 k
1/ 2.2 k  1/ 8.2 k
Eeq
0.2879 V

IL =
= 39.3 μA
Req  RL 1.7346 k  5.6 k
Eeq =
VL = ILRL = (39.3 μA)(5.6 k) = 220 mV
40.
41.
42.
IT = 5 A  0.4 A  0.2 A = 4.40 A
RT = 200   80   50 Ω  50  = 17.39 
VL = ITRT = (4.40 A)(17.39 ) = 75.52 V
V
76.52 V
= 0.38 A
IL = L 
RL
200 
(4 A)(4.7 )  (1.6 A)(3.3 ) 18.8 V + 5.28 V
= 3.01 A

4.7   3.3 
8
Req = 4.7  + 3.3  = 8 
Req ( I eq ) 8 (3.01 A)
=
= 2.25 A
IL =
Req + RL 8  + 2.7 
VL = ILRL = (2.25 A)(2.7 Ω) = 6.08 V
Ieq =

(4 mA)(8.2 k)  (8 mA)(4.7 k)  (10 mA)(2 k)
I eq =
8.2 k  4.7 k  2 k
32.8 V + 37.6 V  20 V
=
= 3.38 mA
14.9 k
Req = 8.2 k + 4.7 k + 2 k = 14.9 k
Req I eq
(14.9 k)(3.38 mA)
IL =
=
= 2.32 mA
Req + RL
14.9 k  6.8 k
VL = ILRL = (2.32 mA)(6.8 k) = 15.78 V
108
CHAPTER 9
43.
15 k  (8 k + 7 k) = 15 k  15 k = 7.5 k
7.5 k(60 V)
= 45 V
Vab =
7.5 k  2.5 k
45 V
Iab =
= 3 mA
15 k
44.
10 V  8 V
2 k  0.51 k  1.5 k
= 498.75 A
V0.51kΩ = (498.75 A)(0.51 k)
= 0.25 V
Vab = 10 V  0.25 V = 9.75 V
Iba =

45.
Vab = 0 V (short)
Iab = 0 A (open)
R2 any resistive value
 R2 = short-circuit, open-circuit, any value
46.
a.
Is =
b.
c.
CHAPTER 9
I
24 V
= 1.5 mA, I = s = 0.5 mA
24 k
3
8 k 
3
24 V
= 0.83 mA
24 k  8 k  12 k
12 k( I s )
I=
= 0.5 mA
12 k  8 k
Is =
yes
109
47.
(a)
10 V
4 k  8 k  4 k  4 k
10 V
=
2.67 k  2 k
10 V
=
= 2.14 mA
4.67 k
IT =
8 ( I T )
= 1.43 mA, I2 = IT/2 = 1.07 mA
8  4
I = I1  I2 = 1.43 mA  1.07 mA = 0.36 mA
I1 =
(8 k  4 k)(10 V)
8 k  4 k  4 k  4 k
= 5.72 V
V
I1 = 1 = 0.71 mA
8 k
V2 = E  V1 = 10 V  5.72 V
= 4.28 V
V2
= 1.07 mA
I2 =
4 k
I = I2  I1 = 1.07 mA  0.71 mA
= 0.36 mA
(b)
48.
a.
b.
110
V1 =
R1 ( I )
3 (6 A)
=2A

R1  R2  R3 3   2   4 
V = I R2 R2 = (2 A)(2 ) = 4 V
I R2 
R2 ( I )
2 (6 A)
= 1.33 A

R1  R2  R3 3   2   4 
V = I R1 R1 = (1.33 A)(3 ) = 4 V
I R1 
CHAPTER 9
Chapter 10
Q1 (9  109 )(4 C)
= 36  103 N/C
=
2
2
r
(1 m)
(a)
E= k
(b)
E = k
2.
E =
kQ
kQ
(9  109 )(2  C)
r
=
=

= 15.81 m
E
72 N/C
r2
3.
C=
Q 1200 C

= 50 μF
V
24 V
4.
Q = CV = (0.15 F)(45 V) = 6.75 μC
5.
a.
1.
b.
6.
7.
8.
Q1
(9  109 )(4  C)
= 36  109 N/C
r
(1 mm) 2
E (1 mm): E (2 m) = 36 × 109: 36 × 103 = 1 × 106

2
 1m 
1" 
 25.4 mm
 39.37" 
V 500 mV
E = =
= 19.69 V/m
d 25.4 mm
25.4 mm
 0.254 mm
100
V
500 mV
E = =
= 1.97 kV/m
d 0.254 mm
Q 160 C

= 23.53 V
C 6.8  F
V 23.53 V
E = =
= 4.71 kV/m
d
5 mm
V=
 1m 
0.1" 
 2.54 mm
 39.37" 
(0.1 m 2 )
A
= 348.43 pF
C = 8.85  1012εr = 8.85  1012(1)
2.54 mm
d
(0.1 m 2 )
A
C = 8.85  1012 εr = 8.85  1012(2.5)
= 871.06 pF
2.54 mm
d
9.
C = 8.85  1012εr
10.
C = εrCo  εr =
CHAPTER 10
A
8.85  1012 (4)(0.15 m 2 )
d=
= 2.66 m
d
2 F
C
6.8 nF
= 5 (mica)
=
Co 1360 pF
111
11.
12.
C = 8.85  10-12(7)
b.
E =
c.
Q = CV = (24.78 nF)(200 V) = 4.96 C
a.
b.
c.
d.
13.
14.
15.
(0.08 m 2 )
= 24.78 nF
0.2 mm
a.
d=
V
200 V

= 106 V/m
d 0.2 mm
1
(4.7 F) = 2.35 F
2
C = 2(4.7 F) = 9.4 F
C = 20(4.7 F) = 94 F
1
(4) 
 3  (4.7 F) = 25.1 F
C=
1
 
 4
C=
8.85  1012  r A (8.85  1012 )(5)(0.02 m 2 )
= 130.15 µm

6800 pF
C
106 m   39.37 in.  1000 mils 
d = 130.15 µm 


 = 5.12 mils
 1  m   1 m   1 in. 
 5000 V 
5.12 mils 
 = 25.6 kV
 mil 
1200 V
 mil 
mica:
 1200 V 
 = 0.24 mils
5000 V
 5000 V 
mil
1

  1m 
0.24 mils 

 = 6.10 m
1000 mils   39.37 in. 
200
(22 F)/C = 4400 pF/C
1  106
4400 pF
4400 pF
[80C] = 0.35 F
[T] 
C
C
16.
J = 5%, Size  40 pF  2 pF, 38 pF  42 pF
17.
F = 1%, Size  47 × 101 = 470 F ± 4.7 F, 465.3 F  474.7 F
18.
K = 10%, Size  18 × 102 pF = 1800 pF  180 pF, 1620 pF  1980 pF
19.
a.
τ = RC = (10
b.
C = E(1  et/τ) = 20 V(1  et/0.56 s)
112
5
Ω)(5.6 μF) = 0.56 s
CHAPTER 10
c.
1τ = 0.632(20 V) = 12.64 V, 3τ = 0.95(20 V) = 19 V
5τ = 0.993(20 V) = 19.87 V
d.
iC =
20 V t/τ
e = 0.2 mAet/0.56 s
100 k
R = Eet/τ = 20 Vet/0.56 s
e.
20.
21.
a.
τ = RC = (10 )(5.6 F) = 5.6 s
b.
υC = E(1  et/τ) = 20 V(1  et/5.6s)
c.
1τ = 12.64 V, 3τ = 19 V, 5τ = 19.87 V
d.
iC =
e.
Same as problem 21 with 5τ = 28 s and Im = 20 A
a.
τ = RC = (2.2 k + 3.3 k)1 μF = (5.5 k)(1 F) = 5.5 ms
b.
C = E(1  et/τ) = 100 V(1  et/5.5 ms)
c.
1τ = 63.21 V, 3τ = 95.02 V, 5τ = 99.33 V
d.
iC =
6
20 V t/τ
e = 20 A et/5.6s
1M
υR = Eet/τ = 20V et/5.6s
E t/τ 100 V t/τ
e =
e = 18.18 mAet/5.5 ms
RT
5.5 k
3.3 k(100 V)
= 60 V
VR2 
3.3 k  2.2 k
 R2 = 60 Vet/5.5 ms
e.
CHAPTER 10
113
22.
a.
R = 68 kΩ + 22 kΩ = 90 kΩ
τ = RC = (90 k)(18 μF) = 1.62 s
b.
C = E(1  et/τ) = (20 V + 40 V)(1  et/τ)
C = 60 V(1  et/1.62s)
c.
iC =
60 V t/τ
E t/τ
e =
e = 0.67 mAet/1.62s
90 k 
R
d.
23.
a.
100 μs
b.
C = 12 V(1  e50µs/100µs) = 12 V(1  e0.5) = 12 V(1  0.607)
c.
C = 12 V(1  e1ms/100µs) = 12 V(1  e10) = 12 V(1  45.4  106)
= 12 V(.393) = 4.72 V
 12 V(999.95 × 103)  12 V
24.
c.
 = 20 ms, 5 = 5(20 ms) = 100 ms
 20 ms
 = RC, R = 
= 2 k
C 10  F
C (20 ms) = 40 mV(1  e20ms/20ms) = 40 mV(1  e1)
d.
e.
f.
C = 40 mV(1  e10) = 40 mV(1  45  106)  40 mV
Q = CV = (10 F)(40 mV) = 0.4 C
 = RC = (1000  106 )(10 F) = 10  103 s
a.
b.
= 40 mV(1  .368) = 40 mV(0.632) = 25.28 mV
1 min   1 h 
5 = 50  103 s 

 = 13.89 h
 60 s   60 min 
25.
a.
 = RC = (4.7 k)(56 F) = 263.2 ms
b.
C = E(1 et/) = 22 V(1  et/263.2ms)
iC =
c.
d.
C(1 s) = 22 V(1  e1s/263.2ms) = 22 V(1  e3.8)
= 22 V(1  22.37  103) = 21.51 V
iC (1 s) = 4.68 mAe1s/263.2ms = 4.68 mA(22.37  103) = 0.105 mA
C = 21.51 Vet/263.2ms
iC =
114
E  t /
22 V  t / 263.2ms
e
e

= 4.68 mAet/263.2ms
R
4.7 k
21.51 V  t / 263.2ms
e
= 4.58 mAet/263.2ms
4.7 k
CHAPTER 10
e.
26.
a.
 = RC = (3 k + 2 k)(2 F) = 10 ms
C = 30 V(1  et/10ms)
30 V  t /10ms
e
= 6 mAt/10ms
5 k
 R1 = iC R1 = (6 mA)(3 k)et/10ms = 18 Vet/10ms
iC =
b.
100ms: e10 = 45.4  106
C = 30 V(1  45.4  106) = 30 V
iC = 6 mA(45.4  106) = 0.27 A
 R1 = 18 V(45.4  106) = 0.82 mV
c.
200 ms:  = R2C = (2 k)(2 F) = 4 ms
C = 30 Vet/4ms
30 V  t / 4ms
e
= 15 mAet/4ms
2 k
At t = 0:  R2  iC R2  (6 mA)(2 k)e  t /10 ms
iC = 
= 12 Vet/10 ms
At t = 200 ms:  R2  (15 mA)(2 k)e t / 4 ms
= 30 Vet/4 ms
d.
CHAPTER 10
115
27.
a.
 = RC = (220 k)(22 pF) = 4.84 s
C = 60 V 1  e  t/4.84  s

iC =
b.

60 V  t / 
e
= 272.73  Ae t/4.84  s
220 k
 = RC = (220 k + 470 kΩ)(22 pF) = (690 kΩ)(22 pF) = 15.18 s
C = 60 V(1  e5) = 60 V(1  6.73 × 103) = 59.6 V
C = 59.6 Vet/15.18μs
iC = 272.73 µAe5 = 272.73 µA(6.73 × 103) = 1.84 μA
60 V
iC(max) =
= 86.96 µA
690 k
iC = 86.96 μAet/15.18μS
c.
116
CHAPTER 10
d.
R = (470 kΩ)(86.96 µA)et/15.18ms
= 40.87Vet/15.18μs
28.
29.
a.
 = RC = (2 m)(1000 F) = 2 s
5 = 10 s
b.
Im =
c.
yes
a.
C = Vf + (Vi  Vf)et/
 = RC = (4.7 k)(4.7 F) = 22.1 ms, Vf = 40 V, Vi = 6 V
C = 40 V + (6 V  40 V)et/22.1ms
C = 40 V  34 Vet/22.1ms
b.
Initially VR = E + C = 40 V  6 V = 34 V
V
34 V  t / 22.1ms
e
iC = R e  t /  
= 7.23 mA et/22.1ms
R
4.7 k
V 12 V

= 6 kA
R 2 m
c.
CHAPTER 10
117
30.
 = RC = (2.2 k)(2000 F) = 4.4 s
C = VCet/ = 40 Vet/4.4 s
VC t / 
40 V  t / 4.4 s
e
e

= 18.18 mAet/4.4 s
R
2.2 k
R = C = 40 Vet/4.4s
IC =
31.
C = Vf + (Vi Vf)et/
 = RC = (820 )(3300 pF) = 2.71 s, Vf = 20 V, Vi = 10 V
C = 20 V + (10 V  (20 V))et/2.71s
C = 20 V + 10 Vet/2.71s
(20 V  10 V) 10 V

= 12.2 mA
820 
820 
iC = iR = 12.2 mAet/2.71s
Im =
32.
118
a.
R = 10 kΩ + 8.2 kΩ = 18.2 kΩ
τ = RC = (18.2 kΩ)(6.8 µF) = 123.76 ms
C = Vf + (Vi  Vf)et/τ
Vf = 20 V + 40 V = 60 V
Vi = 8 V
C = 60 V + (8 V  60 V)et/123.76 ms
C = 60 V  68 Vet/123.76 ms
8 V + 20 V + 40 V
= 3.74 mA
Im =
18.2 k
iC = 3.74 mAet/123.76 ms
CHAPTER 10
b.
33.
a.
C = 140 mV(1  e1ms/2 ms) = 140 mV(1  e0.5) = 140 mV(1  0.6065)
= 140 mV(0.3935) = 55.59 mV
b.
C = 140 mV(1  e10) = 140 mV(1  45.4  106)
 139.99 mV
c.
100 mV = 140 mV(1  et/2 ms)
0.714 = 1  et/2 ms
0.286 = et/2 ms
loge 0.286 = loge et/2 ms
1.252 = t/2 ms
t = 1.252 (2 ms) = 2.5 ms
d.
C = 138 mV = 140 mV(1  et/2 ms)
0.986 = 1  et/2 ms
14 × 103 = et/2 ms
loge 14 × 103 = t/2 ms
4.268 = t/2 ms
t = (4.268)(2 ms) = 8.54 s
CHAPTER 10
119
34.
τ = RC = (33 kΩ)(20 µF) = 0.66 s
C = 12 V(1  et0.66 s)
8 V = 12 V(1  et0.66 s)
8 V = 12 V  12 Ve(1  et/0.66 s)
4 V = 12 Vet0.66 s
0.333 = et0.66 s
loge 0.333 = t/0.66 s
1.0996 = t/0.66 s
t = 1.0996(0.66 s) = 0.73 s
  
t =  loge 1  C 
E 

 12 V 
10 s =  loge 1 

 20 V 
35.
.4
916.29  103
10 s
= 10.92 s
0.916
 10.92 s
 = RC  R = 
= 54.60 k
C 200  F
=
36.
a.
τ = RC = (12 kΩ + 8.2 kΩ)(6.8 µF) = 137.36 ms
C  60 V(1  e  t /  )
48 V  60 V(1  e  t /  )
0.8  1  e  t / 
0.2  e  t / 
log e 0.2  log e e  t / 
b.
c.
1.61  t / 
t  (1.61)  (1.61)(137.36 ms) = 221.15 ms
E
60 V  t / 
iC  e  t /  
e
20.2 k
R
 2.97 mAe  t /137.36 ms
iC(221.15 ms) = 2.97 mAe221.15 ms/137.36 ms
= 2.97 mAe1.61
= 2.97 mA (199.89 × 103)
= 0.594 mA
t = 2τ
iC = 2.97 mAe2τ/τ = 2.97 mAe2
= 0.4 mA
0.135
P = EI = (60 V)(0.4 mA) = 24 mW
120
CHAPTER 10
37.
a.
m = R = Eet/ = 60 Ve1/ = 60 Ve1
= 60 V(0.3679)
= 22.07 V
b.
c.
38.
a.
E t / 
60 V  2 / 
e

e
= 6 Ae2
R
10 M
= 6 A(0.1353)
= 0.81 A
iC =
C = E(1  et/)
t/2 s
50 V = 60 V(1  e )
0.8333 = 1  et/2 s
loge 0.1667 = t/2 s
t = (2 s)(1.792)
= 3.58 s
 = RC = (10 M)(0.2 F) = 2 s
Thevenin’s theorem:
RTh:
ETh:
RTh = 8 k  24 k
= 6 k
ETh = 
24 k(20 V)
= 15 V
24 k  8 k
 = RC = (10 k)(15 F) = 0.15 s
C = E(1  et/)
= 15 V(1  et/0.15 s)
iC =
CHAPTER 10
E  t /
15 V  t / 0.15
e
e

= 1.5 mAet/0.15 s
R
10 k
121
b.
39.
a.
Source conversion and combining series resistors:
E = (4 mA)(6.8 kΩ) = 27.2 V
RT = 6.8 kΩ + 1.5 kΩ = 8.3 kΩ
Vf = 27.2 V, Vi = 10 V
 = RC = (8.3 k)(2.2 F) = 18.26 ms
C = Vf + (Vi  Vf)et/
= 27.2 V + (10 V  (27.2 V))et/18.26 ms
C = 27.2 V + 37.2 Vet/18.26 ms
R(0+) = 27.2 V  (27.2 V))e-t/18.26 ms = 37.2 V
37.2 V  t /18.26ms
iC = 
e
8.3 k
iC = 4.48 mAet/18.26 ms
122
CHAPTER 10
b.
40.
a.
RTh = 3.9 k + 0   1.8 k = 3.9 k
ETh = 36 V
 = RC = (3.9 k)(20 F) = 78 ms
C = Vf + (Vi  Vf)et/
= 36 V + (12 V  36 V)et/78 ms
C = 36 V  24 Vet/78 ms
R(0+) = 24 V  12 V = 24 V
24 V  t/78 ms
e
iC =
3.9 k
iC = 6.15 mAet/78 ms
b.
CHAPTER 10
123
41.
Source conversion:
E = IR1 = (5 mA)(0.56 k) = 2.8 V
R = R1 + R2 = 0.56 k + 3.9 k = 4.46 k
RTh = 4.46 k  6.8 k = 2.69 k
4 V  2 .8 V
1.2 V

= 0.107 mA
I=
6.8 k  4.46 k 11.26 k
ETh = 4 V  (0.107 mA)(6.8 k)
= 4 V  0.727 V
= 3.27 V
C = 3.27 V(1  et/)
 = RC = (2.69 k)(20 F)
= 53.80 ms
C = 3.27 V(1  et/53.80 ms)
3.27 V  t / 
e
2.69 k
= 1.22 mA et/53.80 ms
iC =
42.
a.
 = RC = (6.8 k)(39 F) = 265.2 ms
C = Vf + (Vi  Vf)et/
= 20 V + (8 V  (20 V))et/265.2 ms
C = 20 V + 12 Vet/265.2 ms
R(0 +) = +8 V  20 V = 12 V
12 V  t / 265.2 ms
e
iC = 
6.8 k
iC = 1.76 mAet/265.2 ms
124
CHAPTER 10
b.
43.
a.
 = RThC = (1.67 M)(1 F) = 1.67 s
RTh = 2 M  10 M = 1.67 M
10 M(24 V)
= 20 V
ETh =
10 M  2 M
C = ETh(1  et/)
= 20 V(1  e4/)
= 20 V(1  e4)
= 20 V(1  0.0183)
= 19.63 V
E t / 
e
R
20 V  t /1.67s
3 A =
e
1.67 M
0.25 = et/1.67s
loge 0.25 = t/1.67 s
t = (1.67 s)(1.39)
= 2.32 s
iC =
c.
meter = C
C = ETh(1  et/)
10 V = 20 V(1  et/1.67s)
0.5 = 1  et/1.67s
0.5 = et/1.67s
loge 0.5 = t/1.67 s
t = (1.67 s)(0.69)
= 1.15 s
CHAPTER 10
125
44.
iC ao  C
C
t
(40 V)
= 80 mA
1 ms
(0 V)
1  2 ms: iC = 2  106
= 0 mA
1 ms
(20 V)
2  3 ms: iC = 2  106
= 40 mA
1 ms
(10 V)
= 6.67 mA
3  6 ms: iC = 2  106
3 ms
(0 V)
6 9 ms: iC = 2  106
= 0 mA
3 ms
(10 V)
9 12 ms: iC = 2  106
= 6.67 mA
3 ms
0  1 ms: iC = 2  106
45.
iC ao  C
C
t
(5 V )
= 1.18 A
20  s
(15 V )
20  30 s: iC = 4.7 F
= 7.05 A
10  s
(15 V )
30  40 s: iC = 4.7 F
= 7.05 A
10  s
0  20 s: iC = 4.7 F
(0 V )
=0A
10  s
(5 V )
50  55 s: iC = 4.7 F
= 4.7 A
5 s
40  50 s: iC = 4.7 F
55 s  60 s: iC = 4.7 F
126
(5 V )
= 4.7 A
5 s
CHAPTER 10
60 s  70 s: iC = 4.7 F
(0 V )
=0A
10  s
70 s  80 s: iC = 4.7 F
(10 V )
= 4.7 A
10  s
80 s  100 s: iC = 4.7 F
46.
(5 V )
= 1.175 A
20  s
t
C
 C  (iC )
t
C
0  4 ms: iC = 0 mA C = 0 V
(2 ms)
(40 mA) = 4 V
4  6 ms: iC = 40 mA C =
20 F
(10 ms)
6  16 ms: iC = +40 mA C =
(40 mA) = +20 V
20 F
(4 ms)
16  20 ms: iC = 120 mA C =
(120 mA) = 24 V
20  F
20  25 ms: iC = 0 mA C = 0 V
iC = C
CHAPTER 10
127
47.
6 F + 4 F = 10 µF, 8 µF + 12 µF = 20 µF
10 µF  20 F = 6.67 F
48.
CT = 6 F  12 F = 4 F
CT = CT + 12 F = 4 F + 12 F = 16 F
6 F  CT (6 F)(16 F)
= 4.36 F
CT = 6 F  CT =

6 F  CT
6 F  16 F
V1 = 10 V, Q1 = V1C1 = (10 V)(6 F) = 60 C
CT = 6 F  12 F = 4 F, QT = CTE = (4 F)(10 V) = 40 C
Q2 = Q3 = 40 C
Q
40  C
= 6.67 V
V2 = 2 
C2
6 F
Q
40  C
V3 = 3 
= 3.33 V
C3 12  F
49.
50.
51.
52.
360 µF + 200 µF = 560 µF
470 µF  560 µF = 255.53 µF
QT = Q3 = CTE = (255.53µF)(56 V) = 14.5 mC
Q 14.5 mC
= 30.4 V
V3 = 3 
C3 470  F
V1 = V2 = E  V3 = 56 V  30.4 V = 25.6 V
Q1 = V1C1 = (25.6 V)(360 µF) = 9.2 mC
Q2 = V2C2 = (25.6 V)(200 µF) = 5.1 mC
steady  state  ignore 10 kΩ resistor
330 µF + 120 µF = 450 µF
CT = 220 µF  450 µF = 147.76 µF
QT = Q1 = CTE = (147.76 µF)(20 V) = 2.96 mC
Q 2.96 mC
= 13.45 V
V1 = 1 
C1 220  F
V3 = V2 = E  V1 = 20 V  13.45 V = 6.55 V
Q2 = C2V2 = (330 µF)(6.55 V) = 2.16 mC
Q3 = C3V3 = (120 µF)(6.55 V) = 0.786 mC
4 k(48 V)
= 32 V = V0.08F
4 k  2 k
Q0.08F = (0.08 F)(32 V) = 2.56 C
V0.04F = 48 V
Q0.04F = (0.04 F)(48 V) = 1.92 C
V4k =
53.
WC =
1
1
CV 2  (120 pF)(12 V)2 = 8,640 pJ
2
2
54.
W=
Q2
Q=
2C
128
2CW  2(6 F)(1200 J) = 0.12 C
CHAPTER 10
55.
a.
56.
a.
(220 k  3.3 k)(12 V)
= 9.85 V
2.2 k  3.3 k  1.2 k
(3.3 k)(12 V)
= 5.91 V
V100F =
2.2 k  3.3 k  1.2 k
1
W200F = (200  F)(9.85 V) 2 = 970 mJ
2
1
W100F = (100  F)(5.91 V)2 = 1.75 mJ
2
1
1
WC = CV 2  (1000 F)(100 V)2 = 5 pJ
2
2
V200F =
b.
Q = CV = (1000 F)(100 V) = 0.1 C
c.
I = Q/t = 0.1 C/(1/2000) = 200 A
d.
P = VavIav = W/t = 5 J(1/2000 s) = 10,000 W
e.
t = Q/I = 0.1 C/10 mA = 10 s
CHAPTER 10
129
Chapter 11
1.
a.
b.
c.
d.
2.
3.
 4  104 Wb

= 4  102 Wb/m2 = 0.04 Wb/m2
2
A
0.01 m
0.04 T
F = NI = (40 t)(2.2 A) = 88 At
104 gauss 
3
0.04 T 
 = 0.4  10 gauss
1
T


B=
 2.54 cm   1 m 
= 5.08 mm
0.2 
 1  100 cm 
 2.54 cm   1 m 
= 25.4 mm
1 
 1  100 cm 
 d 2  (5.08 mm) 2

= 20.27  106 m2
A=
4
4
N 2  A (200 t) 2 (4  10 7 )(20.27  10 6 m 2 )
L=
= 40.1 H


25.4 mm
L=
b.
increase = change in µ r
Lnew = µ rLo
4.
L = N2
5.
L=
a.
b.
c.
d.
6.
130
N 2  r o A (200 t ) 2 (500)(4  10 7 )(20.27  10 6 m 2 )
= 20.06 mH


25.4 mm
a.
r o


(200 t) 2 (1000)(4  107 )(1.5  104 m 2 )
= 50.27 mH
0.15 m
N 2  r o A

L = (3)2Lo = 9Lo = 9(4.7 mH) = 42.3 mH
1
1
L = Lo = (4.7 mH) = 1.57 mH
3
3
(2)(2) 2
L =
Lo = 16 (4.7 mH) = 75.2 mH
1
2
2
1 1
  (1500) Lo
2 2
L =  
= 375(4.7 mH) = 1.76 mH
1
2
a.
39  102 H  10%  3900 H  10%  3.9 mH ± 10%
b.
68 × 100 H  5% = 68 F ± 5%
CHAPTER 11
c.
47 μH ± 10%
d.
15 × 102 µH ± 10% = 1500 µH ± 10% = 15 mH ± 10%
7.
e= N
d
= (50 t)(120 mWb/s) = 6.0 V
dt
8.
e= N
d
d e 20 V

 
= 100 mWb/s
dt
dt N 200 t
9.


 1 
d
1


 N = e
= 14 turns
e= N
  42 mV 

dt
 3 m Wb/s 
 d 
 dt 
10.
a.
11.
b.
diL
= (22 mH)(1 A/s) = 22 mV
dt
di
e = L L = (22 mH)(1 mA/ms) = 22 mV
dt
2 mA
diL
e= L
= (22 mH)(
= 4.4 V
10  s
dt
a.
=
b.
iL =
c.
L = Eet/ = 20 Vet/15 s
R = iRR = iLR = E(1  et/) = 20 V(1  et/15 s)
d.
iL: 1 = 0.632 mA, , 3 = 0.951 mA, 5 = 0.993 mA
L: 1 = 7.36 V, 3 = 0.98 V, 5 = 140 mV
e= L
L 300 mH

= 15 s
R
20 k
E
20 mV
(1  e  t/ ) 
(1  et/)
R
20 k
= 1 mA(1  et/15 μs)
e.
CHAPTER 11
131
12.
a.
=
L 4.7 mH

= 2.14 s
R 2.2 k
b.
iL =
E
12 V
(1  e t/ ) 
(1  et/) = 5.45 mA(1  et/2.14 s)
R
2.2 k
c.
L = Eet/ = 12 Vet/2.14 s
R = iRR = iLR = E(1  et/) = 12 V(1  et/2.14 s)
d.
iL: 1 = 3.45 mA, , 3 = 5.18 mA, 5 = 5.41 mA
L: 1 = 4.42 V, 3 = 0.60 V, 5 = 0.08 V
e.
13.
14.
18 V
(1  e  t /  )
RT
18V
= 1.2 kΩ
RT =
15 mA
R (15  s) 1.2 k(15  s)
L

5  15   = 15 µs: L 
R
5
5
 3.6 mH
iL =
a.
iL = If + (Ii  If)et/
E
36 V
L 120 mH


= 9.23 mA,  =
= 30.77 s
Ii = 8 mA, If =
R 3.9 k
R 3.9 k
iL = 9.23 mA + (8 mA  9.23 mA)et/30.77 s
iL = 9.23 mA  1.23 mAet/30.77 s
+E   L   R = 0 and  L  E   R
 R = iRR = iLR = (8 mA)(3.9 k) = 31.2 V
 L = E   R = 36 V  31.2 V = 4.8 V
L = 4.8 Vet/30.77 s
b.
132
CHAPTER 11
15.
a.
Ii = 8 mA, If = 9.23 mA,  =
L 120 mH

= 30.77 s
R 3.9 k
iL = If + (Ii  If)et/
= 9.23 mA + (8 mA  9.23 mA)et/30.77 s
iL = 9.23 mA  17.23 mA et/30.77 s
+E   L   R = 0 (at t = 0)
but,  R = iRR = iLR = (8 mA)(3.9 k) = 31.2 V
 L = E   R = 36 V  (31.2 V) = 67.2 V
L = 67.2 V et/30.77 s
b.
16.
c.
Final levels are the same. Transition period defined by 5 is also the same.
a.
Source conversion:
L
2H

= 588.2 s
R 3.4 k
iL = If + (Ii  If)et/
6V
= 1.76 mA
If =
3.4 k
iL = 1.76 mA + (4 mA  1.76 mA)et/588.2s
iL = 1.76 mA + 2.24 mA et/588.2s
=
R(0 +) = 4 mA(3.4 k) = 13.6 V
KVL: +6 V  13.6 V  L(0+) = 0
L(0+) = 7.6 V
t/588.2s
L = 7.6 Ve
b.
CHAPTER 11
133
17.
a.
20.8 V
= 2 mA
10.4 k
L
200 mH
τ=
=
= 19.23 s
R
10.4 k
iL = If + (Ii  If )et/τ
= 2 mA + (6 mA  2 mA)et/19.23 μs
iL = 2 mA + 4 mAet/19.23 μs
If =
KVL: 20.8 V  62.4 V  υL(0+) = 0
υL(0+) = 41.60 V
υL = 41.6 Ve-t/19.23 μs
b.
18.
a.
=
L = 8 Vet/0.278s, iL =
b.
134
10 mH
L
=
= 0.278 s
36 k
R
E
(1  e t/ ) = 0.222 mA(1  et/0.278s)
R
5  steady state
L
10 mH
 = 
= 0.208 s
R 12 k  36 k
iL = Imet/ = 0.222 mAet/0.208s
L = (0.222 mA)(48 k)et/ = 10 .66Vet/0.308s
CHAPTER 11
19.
a.
b.
L 1 mH

= 0.5 s
R 2 k
E
12 V
iL = (1  e  t /  ) 
(1  e  t /  ) = 6 mA(1  et/0.5s)
R
2 k
L = Eet/ = 12 V et/0.5s
=
iL = 6 mA(1  et/0.5s) = 6 mA(1  e1s/0.5s)
= 6 mA(1  e2) = 5.19 mA
L 1 mH
 = 
= 83.3 ns
iL = Imet/
R 12 k
iL = 5.19 mAet/83.3ns
t = 1 s: L = 12 Vet/0.5s = 12 Ve2 = 12 V(0.1353) = 1.62 V
VL = (5.19 mA)(12 k) = 62.28 V
L = 62.28 Vet/83.3ns
c.
CHAPTER 11
135
20.
a.
RTh = 6.8 k
ETh = 6 V
=
L
5 mH

= 0.74 s
R 6.8 k
E
6V
(1  e t/ ) 
(1  e t/ ) = 0.88 mA(1  et/0.74s)
R
6.8 k
L = Eet/ = 6 Vet/0.74s
iL =
b.
Assume steady state and IL = 0.88 mA
 =
L 5 mH

= 0.33 s
R 15 k
iL = Imet/ = 0.88 mA et/0.33s
L = Vmet/
Vm = ImR = (0.88 mA)(15 k) = 13.23 V
L = 13.23 Vet/0.33s
c.
136
CHAPTER 11
d.
21.
22.
a.
VR2 max = ImR2 = (0.88 mA)(8.2 k) = 7.22 V
RTh = 2 k + 2.2 k + 6.2 k  3 kΩ = 6.22 k
6.2 k(12 V)
= 8.09 V
ETh =
6.2 k  3 k
8.09 V
L 47 mH
If =
= 1.3 mA ,  = 
= 7.56 s
6.22 k
R 6.22 k
iL = 1.3 mA(1  et/7.56s)
L = 8.09 Vet/7.56s
b.
0.632(1.3 mA) = 0.822 mA
0.368(8.09 V) = 2.98 V
a.
Source conversion: E = IR = (4 mA)(12 k) = 48 V, ENet = 48 V  20 V = 28 V
=
iL =
L 2 mH

= 55.56 ns
R 36 k
E
28 V
(1  e  t/ ) 
(1  e  t/ ) = 0.778 mA(1  et/55.56ns)
R
36 k
L = Eet/ = 28 Vet/55.56ns
b.
t = 100 ns:
iL = 0.778 mA(1  e100ns/55.56ns) = 0.778 mA(1  e1.8) = 0.65 mA
0.165
L = 28 Ve1.8 = 4.62 V
CHAPTER 11
137
23.
RTh = 2.2 k  4.7 k = 1.50 k
4.7 k(10 V)
= 6.81 V
ETh = 
4.7 k  2.2 k
L 10 mH
= 
= 6.67 s
R 1.50 k
a.
E
6.81 V
(1  e  t/ )  
(1  e  t/ ) = 4.54 mA(1  et/6.67s)
R
1.5 k
L = Eet/ = EThet/τ = 6.81 Vet /6.67s
iL = 
b.
t = 10 s:
iL = 4.54 mA(1  e10s/6.67s) = 4.54 mA(1  e1.5)
0.223
= 3.53 mA
L = 6.81 V(0.223) = 1.52 V
c.
L 10 mH

= 2.13 s
R 4.7 k
iL = 3.53 mAet/2.13s
At t = 10 s
VL = (3.53 mA)(4.7 k) = 16.59 V
L = 16.59 Vet/2.13s
 =
d.
24.
a.
L = Eet/
=
L
0 .6 H
0 .6 H
= 5 ms


R1  R3 100   20  120 
L = 36 Vet/5 ms
L = 36 Ve25 ms/5 ms = 36 Ve5 = 36 V(0.00674) = 0.24 V
b.
138
L = 36 Ve1 ms/5 ms = 36 Ve0.2 = 36 V(0.819) = 29.47 V
CHAPTER 11
c.


E
(1  e t/ )  R1
 R1  R3

 R1  iR1 R1  iL R1  
 36 V

(1  e t/5ms )  100 
=

120


t/5 ms
))100 
= (300 mA(1  e
5 ms/5 ms
) = 30 V(1  e1)
= 30 V(1  e
= 30 V(1  0.368) = 18.96 V
25.
d.
iL = 300 mA(1  et/5 ms)
100 mA = 300 mA(1  et/5 ms)
0.333 = 1  et/5 ms
0.667 = et/5 ms
loge 0.667 = t/5 ms
0.405 = t/5 ms
t = 0.405(5 ms) = 2.03 ms
e.
None-In parallel with supply.
a.
Ii =
16 V
= 2 mA
4.7 k  3.3 k
t = 0 s: Thevenin:
RTh = 3.3 k + 1 k  4.7 k = 3.3 k + 0.82 k = 4.12 k
1 k(16 V)
= 2.81 V
ETh =
1 k  4.7 k
iL = If + (Ii  If)et/
2.81 V
L
2H
= 0.68 mA,  = 
= 0.49 ms
4.12 k
R 4.12 k
iL = 0.68 mA + (2 mA  0.68 mA)et/0.49 ms
iL = 0.68 mA + 1.32 mAet/0.49 ms
R(0+) = 2 mA(4.12 k) = 8.24 V
KVL(0+): 2.81 V  8.24 V  L = 0
L = 5.43 V
L = 5.43 Vet/0.49 ms
If =
b.
CHAPTER 11
139
26.
a.
8V
= 5.33 mA, VL = 0 V
1.5 k
RTh = (3 k  12 k)  4 k  1.5 k
2.4 k(20 V)
= 7.5 V
ETh =
2.4 k  4 k
Steady-state: I L 
RTh
  1.5 k  1.5 k  0.75 k
ETh
  8 V  7.5 V  15.5 V

L
3 mH

 4 s
R 0.75 k
15.5 V
= 20.67 mA
Ii = 5.33 mA
0.75 k
iL = If + (Ii  If)et/τ
= 20.67 mA + (5.33 mA  20.67 mA)et/4 µs
iL = 20.67 mA  15.34 mAet/4μs
υL = 15.5 Vet/4μs
If =
b.
iL (2 )  20.67 mA  15.34 mA 
e2
0.135
 18.6 mA
 L (2 )  15.5Ve2  15.5V(0.135) = 2.09 V
c.
Ii = 18.6 mA
υL + υR  8 V = 0
υL = 8 V  υR = 8 V  (18.6 mA)(1.5 kΩ)
= 19.9 V
L 3 mH
 
= 2 µs
R 1.5 k
8V
= 5.33 mA
1.5 k
iL = If + (Ii  If)et/τ = 5.33 mA + (18.6 mA  5.33 mA)et/2µs
= 5.33 mA + 13.27 mAet/2µs
υL = 19.9 Vet/2µs
Ii = 18.6 mA
140
If =
CHAPTER 11
27.
a.
RTh = 2 M  10 M = 1.67 M
10 M(24 V)
= 20 V
ETh =
10 M  2 M
ETh
20 V
= 12 μA
=
RTh 1.67 M 
L
5H

 5 s
 
Rmeter 10 M
iL = 12 Aet/5 s
10 A = 12 Aet/5 s
0.833 = et/5 s
loge 0.833 = t/5 s
0.183 = t/5 s
t = 0.183(5 s) = 0.92 s
I L (0 ) =
28.
b.
L (0+) = iL(0+)Rm = (12 A)(10 M) = 120 V
L = 120 Vet/5s = 120 Ve10s/5s = 120 Ve2 = 120 V(0.135) = 16.2 V
c.
L = 120 Ve5/ = 120 Ve5 = 120 V(6.74  103) = 0.81 V
a.
Closed Switch:
RTh = 1.2 k  2.2 k = 0.776 kΩ
1.2 k(24 V)
ETh =
= 8.47 V
1.2 k  2.2 k
CHAPTER 11
141
Open Switch:
RTh
  6.9 k  1.2 k  1.02 k
1.2 k(24 V)
 
ETh
= 3.56 V
8.1 k
3.56 V + υR  υL = 0
υL = 3.56 V + (1.09 mA)(1.02 kΩ)
= 7.57 V
υL = 7.57Vet/1.18 ms
L
1.24

= 1.18 ms
R 1.02 k
3.56 V
Iss =
= 3.49 mA = If
1.02 k
iL  I f  ( I i  I f )e t / 

 3.49 mA+ (  10.91 mA  (3.49 mA)  e t /1.18 ms
iL = 3.49 mA  7.42 mAet/1.18 ms
b.
29.
142
a.
iL = 100 mA(1  e1ms/20ms) = 100 mA(1  e1/20)
= 100 mA(1  e0.05) = 100 mA(1  951.23 103) = 100 mA(48.77  103)
= 4.88 mA
b.
iL = 100 mA(1  e100ms/20ms) = 100 mA(1  e5)
= 99.33 mA
c.
50 mA = 100 mA(1  et/)
0.5 = 1  et/
0.5 = et/
0.5 = et/
loge 0.5 = t/
t = ()(loge 0.5) = (20 ms)(loge 0.5) = (20 ms)(693.15  103)
= 13.86 ms
CHAPTER 11
30.
d.
99 mA = 100 mA(1  et/20 ms)
0.99 = 1  et/20ms
0.01 = et/20ms
0.01 = et/20ms
loge 0.01 = t/20 ms
t = (20 ms)(loge 0.01) = (20 ms)(4.605) = 92.1ms
a.
IL (1τ) = 0.632Imax = 126.4 µA
126.4
Imax =
= 200 μA
0.632
iL  I m (1  e  t /  )
b.
64.4  s


160  A  200  A 1  e  


0.8  1  e
0.2  e
64.4  s

64.4  s

log e 0.2  1.61 

31.
64.4  s

64.4  s
 40  s
1.61
L
L
 40 s =
R
500 
L  20 mH
c.

d.
Im 
a.
L  open circuit equivalent
10 M(16 V)
= 13.33 V
VL =
10 M  2 M
E
 E  (200  A)(500 )  100 mV
R
b.
RTh = 2 M  10 M = 1.67 M
10 M(16 V)
= 13.33 V
ETh =
10 M  2 M
I Lfinal 
CHAPTER 11
ETh 13.33 V
= 7.98 A

RTh 1.67 M
143
iL = 7.98 A(1  et/3 s)
c.
10 A = 7.98 A(1  et/3 s)
1.253 = 1  et/3 s
0.253 = et/3 s
loge(0.253) = t/3s
1.374 = t/3s
t = 1.374(3 s) = 4.12 s
d.
=
L
5H

= 3 s
R 1.67 M
L = 13.33 V et/3 s = 13.33 V e12 s/3 s = 13.33 V e4
= 13.33 V(0.0183) = 0.244 V
32.
eL = L
i
:
t
15 mA 
0  2 ms, eL =  (200 mH) 
= 1.5 V
 2 ms 
 45 mA 
= 0.75 V
2  14 ms, eL = (200 mH) 
 12 ms 
 15 ms 
= 3 V
14  15 ms, eL = (200 mH) 
 1 ms 
15  19 ms, eL = 0 V
 15 mA 
19  22 ms, eL = (200 mH) 
= 1 V
 3 ms 
22  24 ms, eL = 0 V
33.
L = L
i L
t
 15 mA 
= 37.5 mV
0  2 ms: L = (5 mH) 
 2 ms 
 30 mA 
= 37.5 mV
2  6 ms: L = (5 mH) 
 4 ms 
 15 mA 
6  9 ms: L = (5 mH) 
= 25 mV
 3 ms 
9  13 ms: L = 0 V
 5 mA 
13  14 ms: L = (5 mH) 
= 25 mV
 1 ms 
14  17 ms: L = 0 V
144
CHAPTER 11
 5 mA 
17  19 ms: L = (5 mH) 
= 12.5 mV
 2 ms 
34.
L = 10 mH, 4 mA at t = 0 s
t
i
L = L  i   L
t
L
0  5 s: L = 0 V, iL = 0 mA and iL = 4 mA
5 s
(10 V) = 5 mA
5  10 s: iL =
10 mH
2 s
10  12 s: iL =
(+60 V) = +12 mA
10 mH
12  16 s: L = 0 V, iL = 0 mA and iL = 11 mA
8 s
10 V = 8 mA
16  24 s: iL =
10 mH
35.
a.
L3  L4 = 3.5 mH  5.6 mH = 1.953 mH
L2 + L3  L4 = 3.3 mH + 1.953 mH = 5.253 mH
L = L1  (L2 + L3  L4) = 2.4 mH  5.253 mH = 1.647 mH
LT = L4 + L = 9.1 mH + 1.647 mH
LT = 10.75 mH
36.
L2  L3 = 10 mH  30 mH = 7.5 mH
L = L1 + L2  L3 = 47 mH + 7.5 mH = 54.5 mH
L  L4 = 54.5 mH  22 mH = 15.67 mH
CHAPTER 11
145
37.
33 mH + 1.8 mH = 5.1 mH
4.7 mH  5.1 mH = 2.45 mH
38.
LT = 6.2 mH + 12 mH  36 mH + 24 mH = 39.2 mH
CT = 9.1  F + 10 F  91 F = 9.1 F + 9.01 F = 18.11 F
39.2 mH in series with 18.11 F
39.
7 µF  42 µF = 6 µF
12 µF + 6 µF = 18 μF
5 mH + 20 mH = 25 mH
Series combination of 2.2 kΩ resistor, 25 mH coil, 18 µF capacitor
40.
a.
RT = 2 k  8.2 k = 1.61 k, LT = 3 mH  2 mH = 1.2 mH
L  1.2 mH
= T 
= 745.3 µs
RT 1.61 k
E
iL =
(1  e  t /  )
RT
36 V
(1  e  t / 745.3 s ) = 22.36 mA(1  et/745.3µs)
1.61 k
L = Eet/ = 36 Ve t/745.3µ s
=
b.
146
CHAPTER 11
41.
a.
Source conversion: E = 16 V, Rs = 2 k
RTh = 2 k + 2 k  8.2 k = 2 k + 1.61 k = 3.61 k
8.2 k(16 V)
= 12.86 V
ETh =
8.2 k  2 k
E
12.86 V
L 30 mH
= 8.31 s
Im = Th 
 3.56 mA,  = 
RTh 3.61 k
R 3.61 k
iL = 3.56 mA(1  et/8.31s)
 L1   L2 = 12.86 V initially (t = 0+)
L 
10 mH
1
of total = (12.86 V) = 4.29 V
10 mH +20 mH
3
 L = 4.29 Vet/8.31μs
b.
42.
a.
RTh = 10 k  20 k = 6.67 k
20 k(20 V)
= 13.33 V
ETh =
20 k  10 k
LT = 3 H + 4.7 H  10 H = 3 H + 3.197 H = 6.197 H
L
6.197 H
τ= T =
= 0.93 ms
R 6.67 k 
υL = 13.33Vet/0.93 ms
13.33 V
(1  et/τ) = 2 mA(1  et/0.93 ms)
iL =
6.67 k 
CHAPTER 11
147
b.
c.
3.197 H
 0.52  L
3 H + 3.197 H
 L3 = (0.52)(13.33et/0.93 ms) = 6.93 Vet/0.93 ms
 L3 =
43.
44.
45.
E 20 V
=5A

R1
4
E
20 V
20 V
I2 = I R2 
=2A


R2  R3 6   4  10 
I1 = I R1 + I2 = 5 A + 2 A = 7 A
I R1 
I1 = I2 = 0 A
V1 = V2 = E = 60 V
12 V
= 3 A, I2 = 0 A
4
V1 = 12 V, V2 = 0 V
I1 =
46.
6  V)
= 10.34 V
6   20   3 
(3   6 )(50 V)
= 15.52 V
V1 =
29 
50 V
I1 =
= 1.72 A
20   3   6 
I2 = 0 A
V2 =
148
CHAPTER 11
Chapter 12
1.
Φ: CGS: 5  104 Maxwells, English: 5  102 4 lines
B: CGS: 8 Gauss, English: 51.62 lines/in.
2.
Φ: SI 6  104 Wb, English 60,000 lines
B: SI 0.465 T, CGS 4.65  103 Gauss, English 30,000 lines/in.2
3.
a.
B=
 4  104 Wb
=
= 0.04 T
A
0.01 m 2
4.
a.
R=
0.06 m
300
l
=
=
4
2
 A  2  10 m
m
b.
R=
l
0.0762 m
152.4
=
=
4
2
 A  5  10 m
m
c.
R=
l
0.1 m
1000
=
=
4
2
 A 1  10 m
m
from the above R (c) > R (a) > R (b)
5.
R=
F
400 At
=
= 952.4  103 At/Wb
 4.2  104 Wb
6.
R=
F
120 gilberts
=
= 1.67  103 rels (CGS)
 72,000 maxwells
7.
8.
9.
 1m 
= 0.1524 m
6 in. 
 39.37 in . 
F
400 At
= 2624.67 At/m
H= =
l 0.1524 m
μ=
2 B 2(1200  104 T)
= 4  104 Wb/Am
=
H
600 At/m
 10  104 Wb
=
= 0.33 T
A
3  103 m 2
Fig. 12.7: H  800 At/m
NI = Hl  I = Hl/N = (800 At/m)(0.2 m)/75 t = 2.13 A
B=
CHAPTER 12
149
10.
11.
12.
 3  104 Wb
=
= 0.6 T
A 5  104 m 2
Fig. 12.7, Hiron = 2500 At/m
Fig. 12.8, Hsteel = 70 At/m
NI = Hl(iron) + Hl(steel)
(100 t)I = (Hiron + Hsteel)l
(100 t)I = (2500 At/m + 70 At/m)0.3 m
771 A
= 7.71 A
I=
100
B=
a.
N1I1 + N2I2 = Hl
 12  104 Wb
B= =
=1T
A 12  104 m 2
Fig. 12.7: H  750 At/m
N1(2 A) + 30 At = (750 At/m)(0.2 m)
N1 = 60 t
b.
μ=
a.
B
1T
= 13.34  104 Wb/Am
=
H 750 At/m
 1 Wb 
= 8  104  108 Wb = 8  104 Wb
80,000 lines  8

10 lines 
 1m 
= 0.14 m
l(cast steel) = 5.5 in. 
 39.37 in . 
 1m 
l(sheet steel) = 0.5 in. 
= 0.013 m
 39.37 in . 
 1m   1m 
Area = 1 in.2 
= 6.45  104 m2



39.37
in
.
39.37
in.




8  104 Wb
=
= 1.24 T
A 6.45  104 m 2
Fig 12.8: Hsheet steel  460 At/m, Fig. 12.7: Hcast steel  1275 At/m
NI = Hl(sheet steel) + Hl(cast iron)
= (460 At/m)(0.013 m) + (1275 At/m)(0.14 m)
= 5.98 At + 178.50 At
NI = 184.48 At
B=
b.
150
B
1.24 T
= 9.73  104 Wb/Am
=
H 1275 At/m
B
1.24 T
Sheet steel: μ =
= 26.96  104 Wb/Am
=
H 460 At/m
Cast steel: μ =
CHAPTER 12
13.
N1I + N2 =
+ Hl
Hl


cast steel cast iron
(20 t)I + (30 t)I = "
(50 t)I = "
B=

 1m   1m 
with 0.25 in.2 
= 1.6  104 m2



A
 39.37 in .   39.37 in. 
0.8  104 Wb
= 0.5 T
1.6  104 m 2
Fig. 12.8: Hcast steel  280 At/m
Fig. 12.7: Hcast iron  1500 At/m
 1m 
lcast steel = 5.5 in. 
= 0.14 m
 39.37 in . 
B=
 1m 
= 0.064 m
lcast iron = 2.5 in. 
 39.37 in . 
(50 t)I = (280 At/m)(0.14 m) + (1500 At/m)(0.064 m)
50I = 39.20 + 96.00 = 135.20
I = 2.70 A
14.
15.
a.
lab = lef = 0.05 m, laf = 0.02 m, lbc = lde = 0.0085 m
NI = 2Hablab + 2Hbclbc + Hfalfa + Hglg
 2.4  104 Wb
= 1.2 T  H  360 At/m (Fig. 12.8)
B= =
A
2  104 m 2
100I = 2(360 At/m)(0.05 m) + 2(360 At/m)(0.0085 m)
+ (360 At/m)(0.02 m) + 7.97  105(1.2 T)(0.003 m)
= 36 At + 6.12 At + 7.2 At + 2869 At
100I = 2918.32 At
I  29.18 A
b.
air gap: metal = 2869 At:49.72 At = 58.17:1
B
1.2 T
= 3.33  103 Wb/Am
=
μsheet steel =
H 360 At/m
μair = 4π  107 Wb/Am
μsheet steel: μair = 3.33  103 Wb/Am:4  107  2627:1
 1m 
4 cm 
= 0.04 m
100 cm 
(8  104 Wb  0.5  104 Wb) 36(7.5  104 )
1
d 1
f = NI
= (80 t)(0.9 A)
=
1
0.02
2
dx 2
(0.04 m)
2
= 1.35 N
CHAPTER 12
151
16.
C = 2πr = (6.28)(0.3 m) = 1.88 m
 2  104 Wb
= 1.54 T
B= =
A 1.3  104 m 2
Fig. 12.7: Hsheet steel  2100 At/m
Hg = 7.97  105Bg = (7.97  105)(1.54 T) = 1.23  106 At/m
N1I1 + N2I2 = Hglg + Hl(sheet steel)
(200 t)I1 + (40 t)(0.3 A) = (1.23  106 At/m)(2 mm) + (2100 At/m)(1.88 m)
I1 = 31.98 A
17.
a.
 1m 
= 2  103 m
0.2 cm 
100 cm 
 d 2 (3.14)(0.01 m) 2
= 0.79  104 m2
A=
=
4
4
NI = Hglg, Hg = 7.96  105 Bg

 0.2  104 Wb  
2  103 m
(200 t)I = (7.96  105 ) 
4
2 

 0.79  10 m  
I = 2.02 A
b.
18.

2  104 Wb
=
= 0.25 T
A 0.79  104 m 2
2
1 Bg A 1 (0.25 T ) 2(0.79  104 m 2 )
F
=
2 o
2
4  107
 2N
Bg =
Table:
Section
Φ(Wb)
ab, gh
4
bc, fg
2  10
cd, ef
2  104
5  10
4
5  104
5  10
bg
152
5  10
4
4
ah
de
A(m2)
2  10
4
2  10
4
5  10
4
B(T)
H
l(m)
Hl
0.2
0.1
0.099
0.2
0.2
0.002
CHAPTER 12
 2  104 Wb
=
= 0.4 T
A 5  104 m 2
Air gap: Hg = 7.97  105(0.4 T) = 3.19  105 At/m
Hglg = (3.19  105 At/m)(2 mm) = 638 At
Fig 12.8: Hbc = Hcd = Hef = Hfg = 55 At/m
Hbclbc = Hfglfg = (55 At/m)(0.1 m) = 5.5 At
Hcdlcd = Heflef = (55 At/m)(0.099 m) = 5.45 At
Bbc = Bcd = Bg = Bef = Bfg =
For loop 2:  F = 0
Hbclbc + Hcdlcd + Hglg + Heflef + Hfglfg  Hgblgb = 0
5.5 At + 5.45 At + 638 At + 5.45 At + 5.50 At  Hgblgb = 0
Hgblgb = 659.90 At
659.90 At
= 3300 At/m
and Hgb =
0.2 m
Fig 12.7: Bgb  1.55 T
with Φ2 = BgbA = (1.55 T)(2  104 m2) = 3.1  104 Wb
ΦT = Φ1 + Φ2
= 2  104 Wb + 3.1  104 Wb
= 5.1  104 Wb = Φab = Φha = Φgh

5.1  104 Wb
= 1.02 T
Bab = Bha = Bgh = T =
A
5  104 m 2
BH curve: (Fig 12.8):
Hab = Hha = Hgh  180 At/m
Hablab = (180 At/m)(0.2 m) = 36 At
Hhalha = (180 At/m)(0.2 m) = 36 At
Hghlgh = (180 At/m)(0.2 m) = 36 At
which completes the table!
Loop #1:  F = 0
NI = Hablab + Hbglbg + Hghlgh + Hahlah
(200 t)I = 36 At + 659.49 At + 36 At + 36 At
(200 t)I = 767.49 At
I  3.84 A
19.
NI = Hl
l = 2πr = (6.28)(0.08 m) = 0.50 m
(100 t)(2 A) = H(0.50 m)
H = 400 At/m
Fig. 12.8: B  0.68 T
Φ = BA = (0.68 T)(0.009 m2)
Φ = 6.12 mWb
CHAPTER 12
153
20.
NI = Hab(lab + lbc + lde + lef + lfa) + Hglg
300 At = Hab(0.8 m) + 7.97  105 Bg(0.8 mm)
300 At = Hab(0.8 m) + 637.6 Bg
Assuming 637.6 Bg  Hab(0.8 m)
then 300 At = 637.6 Bg
and Bg = 0.47 T
Φ = BA = (0.47 T)(2  104 m2) = 0.94  104 Wb
Bab = Bg = 0.47 T  H  270 At/m (Fig. 12.8)
300 At = (270 At/m)(0.8 m) + 637.6(0.47 T)
300 At  515.67 At
 Poor approximation!
300 At
 100%  58%
515.67 At
Reduce Φ to 58%
0.58(0.94  104 Wb) = 0.55  104 Wb
 0.55  104 Wb
B= =
= 0.28 T  H  190 At/m (Fig. 12.8)
A
2  104 m 2
300 At = (190 At/m)(0.8 m) + 637.6(0.28 T)
300 At  330.53 At
Reduce Φ another 10% = 0.55  104 Wb  0.1(0.55  104 Wb)
= 0.495  104 Wb
 0.495  104 Wb
= 0.25 T  H  175 At/m (Fig. 12.7)
B= =
A
2  104 m 2
300 At = (175 At/m)(0.8) + 637.6(0.28 T)
300 At  318.53 At but within 5%  OK
Φ  0.55  104 Wb
21.
a.
1τ = 0.632 Tmax
Tmax  1.5 T for cast steel
0.632(1.5 T) = 0.945 T
At 0.945 T, H  700 At/m (Fig. 12.7)
 B = 1.5 T(1  eH/700 At/m)
b.
H = 900 At/m:
B = 1.5 T 1  e

900 At/m
700 At/m
Graph:  1.1 T
H = 1800 At/m:


 = 1.09 T
1800 At/m

B = 1.5 T 1  e 700 At/m = 1.39 T
Graph:  1.38 T
H = 2700 At/m:
B = 1.5 1  e

2700 At/m
700 At/m
 = 1.47 T
Graph:  1.47 T
Excellent comparison!
154
CHAPTER 12
c.
B = 1.5 T(1  eH/700 At/m) = 1.5 T  1.5 TeH/700 At/m
B  1.5 T = 1.5 TeH/700 At/m
1.5  B = 1.5 TeH/700 At/m
1.5 T  B
= eH/700 At/m
1.5 T
B 
H

loge 1 
=
 1.5 T  700 At/m
B 

and H = 700 loge 1 

 1.5 T 
d.
B = 1 T:
1T 

H = 700 loge 1 
 = 769.03 At/m
 1.5 T 
Graph:  750 At/m
B = 1.4 T:
 1.4 T 
H = 700 loge 1 
 = 1895.64 At/m
 1.5 T 
Graph:  1920 At/m
e.
B 

H = 700 loge 1 

 1.5 T 
 0.2 T 
= 700 loge 1 

 1.5 T 
= 100.2 At/m
Hl (100.2 At/m)(0.16 m)
I=
= 40.1 mA
=
N
400 t
vs 44 mA for Ex. 12.1
CHAPTER 12
155
Chapter 13
1.
a.
b.
c.
d.
e.
10 V
15 ms: 10 V, 20 ms: 0 V
20 V
20 ms
2 cycles
2.
a.
b.
c.
d.
e.
200 μA
1 s: 200 μA, 7 s: 200 µA
400 μA
4 s
2.5 cycles
3.
a.
b.
c.
d.
e.
40 mV
1.5 ms: 40 mV, 5:1 ms: 40 mV
80 mV
2 ms
3.5 cycles
4.
a.
T=
b.
c.
d.
5.
a.
b.
c.
d.
1
1
= 5 ms

f 200 Hz
1
1
T= 
= 25 ns
f 40 MHz
1
1
T= 
= 50 s
f 20 kHz
1
1
=1s
T=

f 1 Hz
1 1

= 1 Hz
T 1s
1
1
f= 
= 16 Hz
1
T
s
16
1
1
f= 
= 25 Hz
T 40 ms
1
1
= 40 kHz
f= 
T 25 s
f=
6.
T=
1
= 1 ms, 5(1 ms) = 5 ms
1 kHz
7.
T=
24 ms
= 0.3 ms
80 cycles
156
CHAPTER 13
8.
f=
42 cycles
= 7 Hz
6s
9.
a.
Vpeak = (2.5 div.)(50 mV/div) = 125 mV
b.
T = (3.2 div.)(10 s/div.) = 32 s
c.
f=
a.
  
Radians = 
40 = 0.22 π rad
 180 
10.
b.
c.
11.
  
Radians = 
170 = 0.94 rad
 180 
a.
 180  
= 60
Degrees = 
   3
b.
 180 
Degrees = 
1.2 = 216
  
d.
a.
b.
c.
d.
13.

  
Radians = 
60 = rad
3
 180 
  
Radians = 
135 = 0.75 rad
 180 
d.
c.
12.
1
1
= 31.25 kHz

T 32  s
a.
b.
c.
d.
 180  1
Degrees = 
  = 18
   10
 180 
Degrees = 
 0.6  = 108
  
2
2

= 3.49 rad/s
T 1.8 s
2
=
= 20.94  103 rad/s
3
0.3  10 s
2
=
= 785.4  103 rad/s
8  106 s
2
=
= 1.57 × 106 rad/s
6
4  10 s
=
 = 2 f = 2 (100 Hz) = 628.32 rad/s
 = 2 f = 2 (0.25 kHz) = 1.57 × 103 rad/s
 = 2 f = 2 (2 kHz) = 12.56  103 rad/s
 = 2 f = 2 (0.004 MHz) = 25.13  103 rad/s
CHAPTER 13
157
14.
15.
a.
2

f=
T
2
2 1
T=


f
 754 rad/s

= 120 Hz, T = 8.33 ms
f=
2
2
 = 2 f =
b.
f=
 12 rad/s

= 1.91 Hz, T = 523.6 ms
2
2
c.
f=
 6000 rad/s

= 954.93 Hz, T = 1.05 ms
2
2
d.
f=
 0.16 rad/s

= 25.46  103 Hz, T = 39.28 ms
2
2
   
radians
(60) 

 180  3
t=
  / 3 rad
 / 3 rad
1
1




= 2.78 ms

2 f
2 (60 Hz) (6)(60) 360
16.

 /6
   
= 104.7 rad/s
(30) 
  ,  = t   = 
t 5  10 3 s
 180  6
17.
a.
Amplitude = 20, f =
b.
Amplitude = 12, f = 120 Hz
 10,000 rad/s

= 1591.55 Hz
Amplitude = 106, f =
2
2
 10,058 rad/s

= 1.6 kHz
Amplitude = 8, f =
2
2
c.
d.
 377 rad/s

= 60 Hz
2
2
18.

19.

20.
T=
21.
i = 0.5 sin 72 = 0.5(0.9511) = 0.48 A
22.
158
2


2
1
= 40 ms, cycle = 20 ms
157
2
 180 
1.2 
 = 216
  
 = 20 sin 216 = 20(0.588) = 11.76 V
CHAPTER 13
23.
6  103 = 30  103 sin 
0.2 = sin 
 = sin1 0.2 = 11.54 and 180  11.54 = 168.46
24.
 = Vm sin 
30 1 ms

360
T
 360 
T = 1 ms 
 = 12 ms
 30 
1
1
= 83.33 Hz
f= 
T 12  10 3 s
 = 2 f = (2)(83.33 Hz) = 523.58 rad/s
40 = Vm sin 30 = Vm (0.5)
40
= 80 V
Vm =
0 .5
and  = 80 sin 523.58t
25.

26.

27.
a.
 = 6 × 103 sin (2π 2000t + 30)
b.
i = 20  103 sin(2π 60t  60)
28.
a.
 = 120  106 sin(2π 1000t  80)
29.
 = 12  103 sin(2π 2000t + 135°)
30.
 = 8  103 sin(2π 500t +π/6)
31.
 leads i by 90
32.
i leads  by 40
33.
 = 2 sin (t  30 + 90)
i = 5 sin(t + 60)
+60
in phase
34.
 = 4 sin(t + 90 + 90 + 180 = 4 sint
i = sin(t + 10 + 180) = sin(t + 190)
35.
T=
1
1
= 1 ms

f 1000 Hz
t1 =
120  T  2  1 ms  1
  
 = ms
180  2  3  2 
3
CHAPTER 13
i leads  by 190
159
36.
  2 f 
T
2


2
T
2
 125.66  s
50,000 rad/s
40
40
(T ) 
(125.66  s)
360
360
 13.96  s
t1 
37.
T = 1 ms
tpeak @ 30°
30
1
tpeak =
(T )  ms
360
12
38.
a.
T = ( 8 div.)(1 ms/div.) = 8 ms (both waveforms)
b.
f=
c.
Peak = (2.5 div)(0.5 V/div.) = 1.25 V
Vrms = 0.707(1.25 V) = 0.884 V
d.
Phase shift = 4.6 div., T = 8 div.
4.6 div.
=
 360 = 207 i leads e
8 div.
or e leads i by 153
39.
40.
1
1
= 125 Hz (both)

T 8 ms
0  (6 V)(5 ms)  (3 V)(10 ms)  (3 V)(10 ms)
30 ms
30 V +30 V  30 V

 1V
30
G
1
1

 2 (4 ms)(20 mA)   (2 ms)(8 mA)  2 (2 ms)(8 mA)
G
8 ms
40 mA  16 mA  8 mA 16 mA


8
8
= 2 mA
(35 V)(5 ms) 
41.
G
1
2
(20 V)(20 ms)  (20 V)(15 ms) 
1
(20 V)(7.5 ms) 
2
75 ms
1
2
(20 V)(7.5 ms) 
1
(20 V)(15 ms) + 0
2
175 V  200 V 300 V  75 V  75 V 150 V
75
400 V  575 V

 2.33 V
75

160
CHAPTER 13
42.
43.
1
1
0  (30 mA)(3 ms)  (20mA)(2 ms)
2
2
G
7 ms
45 mA  20 mA

 3.57 mA
7
a.
b.
c.
44.
45.
1
1
(4 V)(5 ms)  (8 V)(5 ms)  (8 V)(5 ms)  (4 V)(5 ms)  (8 V)(5 ms)  (8 V)(5 ms)
2
2
G
25 ms
20 V  20 V  40 V  20 V  20 V  40 V

25
= 0V
The same
1
1
( r 2 )  ( 202 )  628.32
2
2
628.32 628.32

 15.71  15.71 mA
Area =
d
40
(15.71mA)( )  (5 mA)( )
G
2
 5.36 mA
Area =
a.
c.
T = ( 2 div.)(0.2 ms/div) = 0.4 ms
1
1
= 2.5 kHz
f= 
T 0.4 ms
Average = (2.5 div.)(10 mV/div.) = 25 mV
a.
T = (4 div.)(10 s/div.) = 40 s
b.
f=
c.
G=
b.
46.
0V
1
1
= 25 kHz

T 40 s
( 2.5 div.)(1.5 div.)  (1 div.)(0.5 div.)  (1 div.)(0.6 div.)  (2.5 div.)(0.4 div.)(1 div.)(1 div.)
4 div.
3.75 div.  0.5 div.  0.6 div.  1 div.  1 div.
4
6.85 div.
=
= 1.713 div.
4
1.713 div.(10 mV/div.) = 17.13 mV
=
47.
a.
b.
c.
Vrms = 0.7071(120 V) = 84.85 V
Irms = 0.7071(6 mA) = 4.24 mA
Vrms = 0.7071(8 V) = 5.66 V
CHAPTER 13
161
48.
a.
b.
c.
 = 6.79 sin 377t
i = 70.7  103 sin 377t
 = 2.83  103 sin 377t
49.
Vrms =
50.
Vrms =
=
51.
G=
a.
(3 V) 2 (2 s)  (2 V)2 (2 s)  0  (1 V) 2 (2 s)  (  3 V) 2 (2 s)  (  2 V)2 (2 s)
12 s
54 2
V  4.5 V 2 = 2.12 V
12
(8 V)(4 ms)  (8 V)(4 ms)
0
=0V

8 ms
8 ms
Vrms =
52.
1 
( 2 V)2 (4 s)  (2 V)2 (1 s)  (3 V)2  s 
2 
= 1.43 V
12 s
(8 V) 2 (4 ms)  (  8 V) 2 (4 ms)
=8V
8 ms
T = (4 div.)(10 s/div.) = 40 s
1
1
= 25 kHz
f= 
T 40 s
Av. = (1 div.)(20 mV/div.) = 20 mV
Peak = (2 div.)(20 mV/div.) = 40 mV
rms =
b.
162
2
(40 mV)2
Vmax
 (20 mV)2 
= 34.64 mV
2
2
T = (2 div.)(50 s) = 100 s
1
1
= 10 kHz
f= 
T 100 s
Av. = (1.5 div.)(0.2 V/div.) = 0.3 V
Peak = (1.5 div.)(0.2 V/div.) = 0.3 mV
rms =
53.
V02 
V02 
2
(.3 V)2
Vmax
 (.3 V)2 
= 367.42 mV
2
2
a.
CHAPTER 13
1
1
(2)(16)  (2)(16)  (2)(48)  96
2
2
Area = 96 + (4)(64) + (2)(4) = 96 + 256 + 8 = 360
b.
c.
d.
54.
A1 =
rms =
360
 30 = 5.48
12
1
(4)(8)  4(8)  2(2)
16  32  4
G= 2
= 3.67

12
12
e.
rms  1.5 (average value)
a.
Vdc = IR = (4 mA)(2 k) = 8 V
Meter indication = 2.22(8 V) = 17.76 V
b.
Vrms = 0.707(16 V) = 11.31 V
CHAPTER 13
163
Chapter 14
1.

2.

3.
a.
(377)(10) cos 377t = 3770 cos 377t
b.
(200)(0.6) cos(754t + 20) = 120 cos(754t + 20)
c.
( 2 20)(157) cos(157t  20) = 4440.63 cos(157t  20)
d.
(200)(1) cos(t + 180) = 200 cos(t + 180) = 200 cos t
a.
Im = Vm/R = 150 V/3 Ω = 50 A, i = 50 sin 200t
b.
Im = Vm/R = 30 V/3 Ω = 10 A, i = 10 sin(377t + 20)
c.
Im = Vm/R = 6 V/3 Ω = 2 A, i = 2 sin(ωt + 100)
d.
Im = Vm/R = 12 V/3 Ω = 4 A, i = 4 sin(ωt + 220)
a.
Vm = ImR = (0.1 A)(7  103 Ω) = 700 V
υ = 700 sin 1000t
b.
Vm = ImR = (2  103 A)((7  103 Ω) = 14.8 V
υ = 14.8 sin(400t  120)
a.
0
b.
XL = 12.56f = 12.56(60 Hz) = 753.6 
c.
XL = 12.56f = 12.56(4 kHz) = 50.24 k
d.
XL = 12.56f = 12.56(1.2 MHz) = 15.07 M
a.
L=
XL
2 k
=
= 22 mH
2 f 2 (14.47 kHz)
b.
L=
XL
40 k
=
= 1.2 H
2 f 2 (5.3 kHz)
a.
XL = 2πfL  f =
4.
5.
6.
7.
8.
f=
164
XL
XL
XL
=
=
2 L (6.28)(1 mH) 6.28  10 3 H
10 
= 1.59 kHz
6.28  103 H
CHAPTER 14
9.
10.
b.
f=
4 k
XL
= 636.94 kHz

3
6.28  10 H 6.28  10 3 H
c.
f=
12 k
XL
= 1.91 MHz

3
6.28  10 H 6.28  10 3 H
a.
Vm = ImXL = (5 A)(20 ) = 100 V
υ = 100 sin(ωt + 90)
b.
Vm = ImXL = (40  103 A)(20 ) = 0.8 V
υ = 0.8 sin(ωt + 150)
c.
i = 6 sin(ωt + 150), Vm = ImXL = (6 A)(20 ) = 120 V
υ = 120 sin(ωt + 240) = 120 sin(ωt  120)
a.
b.
11.
12.
XL = ωL = (400 rad/s)(0.1 H) = 40 
Vm = ImXL = (5  106 A)(40 ) = 200 μV
υ = 200  106 sin(400t + 110)
a.
Im =
Vm 120 V
= 2.4 A, i = 2.4 sin(ωt  90)
=
X L 50 
b.
Im =
Vm 30 V
= 0.6 A, i = 0.6 sin(ωt  70)
=
X L 50 
a.
b.
13.
XL = ωL = (100 rad/s)(0.1 H) = 10 
Vm = ImXL = (10 A)(10 ) = 100 V
υ = 100 sin(100t + 90)
XL = ωL = (60 rad/s)(0.2 H) = 12 
Im = Vm/XL = 1.5 V/12  = 0.125 A
i = 0.125 sin(60t  90)
XL = ωL = (10 rad/s)(0.2 H) = 2 
Im = Vm/XL = 16 mV/2  = 8 mA
i = 8  103 sin(t + 2  90) = 8  103 sin(t  88)
a.
XC =
1
1
=
=
2 fC 2 (0 Hz)(5  10 6 F)
b.
XC =
1
1
= 530.79 
=
2 fC 2 (60 Hz)(5  106 F)
c.
XC =
1
1
= 15.92 
=
2 fC 2 (2 kHz)(5  106 F)
CHAPTER 14
165
d.
14.
15.
16.
17.
XC =
XC 
1
1
C 
2 fC
2 fX C
a.
C
b.
C
a.
f=
1
1
= 4.08 kHz
=
2 CX C 2 (3.9  10 6 F)(10 )
b.
f=
1
1
= 34 Hz
=
2 CX C 2 (3.9  10 6 F)(1.2 k)
c.
f=
1
1
= 408.1 kHz
=
2 CX C 2π(3.9  10 6 F)(0.1 )
d.
f=
1
1
= 20.40 Hz
=
6
2 CX C 2 (3.9  10 F)(2000 )
1
2 (265 Hz)(60 )
 10  F
1
2 (34 kHz)(1.2 k)
 3900 pF
Im = Vm/XC = 120 V/2.5  = 48 A
i = 48 sin(ωt + 90)
a.
b.
Im = Vm/XC = 4 × 103 V/2.5  = 0.16 A
i = 1.6 × 103 sin(ωt + 130)
a.
υ = 30 sin 200t, XC =
Im =
b.
1
1
= 5 k
=
ωC (200)(1  106 F)
Vm 30 V
= 6 mA, i = 6  103 sin(200t + 90)
=
X C 5 k
υ = 60  103 sin 377t, XC =
Im =
166
1
1
= 62.83 
=
6
2 fC 2 (2  10 Hz)(5  106 F)
1
1
= 2.65 k
=
ωC (377)(1  106 )
3
V m 60  10 V
=
= 22.64 A, i = 22.64  106 sin(377t + 90)

2,650
XC
CHAPTER 14
18.
Vm = ImXC = (50  103 A)(10 ) = 0.5 V
υ = 0.5 sin(ωt  90)
a.
19.
b.
Vm = ImXC = (2  106)(10 ) = 20 V
υ = 20  106 sin(ωt  30)
a.
i = 0.2 sin 300t, XC =
1
1
= 5.952 k

ωC (300)(0.56  106 F)
Vm = ImXC = (0.2 A)(5.952 kΩ) = 1190.48 V, υ = 1190.48 sin(300t  90)
b.
i = 8  103 sin (377t 30°), XC =
1
1
= 4.737 k

ωC (377)(0.56  106 F)
Vm = ImXC = (8  103 A)(4.737 k) = 37.81 V
υ = 37.81 sin(377t  120)
20.
21.
a.
υ leads i by 90  L, XL = Vm/Im = 550 V/11 A = 50 
X
50 
= 132.63 mH
L= L=
ω 377 rad/s
b.
υ leads i by 90  L, XL = Vm/Im = 36 V/4 A = 9 
1
1
= 147.36 μH
L=
=
ωX L (754 rad/s)(9 )
c.
υ and i are in phase  R
V
10.5 V
=7
R= m=
I m 1.5 A
a.
b.
c.
22.
i = 5 sin(ωt + 90) 
 i leads  by 90  C
 = 2000 sin ωt

Vm 2000 V
= 400 
XC =
=
5A
Im
i = 2 sin(157t + 60)

  leads i by 90  L
 = 80 sin(157t + 150) 
V
80 V
40 
XL = m =
= 40 Ω, L = X L =
= 254.78 mH
ω 157 rad/s
Im 2 A
 = 35 sin(ωt  20) 
 in phase  R
i = 7 sin(ωt  20)

Vm 35 V
R=
=5
=
Im 7 A

CHAPTER 14
167
23.
24.

XC =
1
1
1
1
=R  f =
=
=
3
6
2 fC
2 RC 2 (2  10 )(1  10 F) 12.56  103
 79.62 Hz
25.
XL = 2πfL = R
R
10,000 
= 318.47 mH
L=
=
2 f 2 (5  103 Hz)
26.
XC = XL
1
 2 fL
2 fC
1
f2 =
4 2 LC
1
1
=
= 1.59 kHz
and f =

3
2 LC 2 (10  10 H)(1  106 F)
27.
XC = XL
1
1
1
= 2 fL  C = 2 2 =
= 5.07 nF
2 fC
4 f L 4(9.86)(2500  106 )(2  103 )
28.
a.
P=
(60 V)(15 A)
cos 30 = 389.7 W, Fp = 0.866
2
b.
P=
(50 V)(2 A)
cos 0 = 50 W, Fp = 1.0
2
c.
P=
(50 V)(3 A)
cos 10 = 73.86 W, Fp = 0.985
2
d.
P=
(75 V)(0.08 A)
cos 40 = 2.30 W, Fp = 0.766
2
2
29.
168
8A
V m 48 V
= 6 , P = I2R = 
=
 6  = 192 W
Im 8A
 2
V I
(48 V)(8 A)
P = m m cos  =
cos 0 = 192 W
2
2
 48 V  8 A 
P = VI cos θ = 

 cos 0 = 192 W
 2  2 
All the same!
R=
CHAPTER 14
30.
P = 100 W: Fp = cos θ = P/VI = 100 W/(150 V)(2 A) = 0.333
P = 0 W: Fp = cos θ = 0
300
=1
P = 300 W: Fp =
300
31.
P = V m I m cos 
2
(50 V) I m
(0.5)  Im = 40 A
500 W =
2
i = 40 sin(ωt  50)
32.
a.
Im = Em/R = 34 V/6.8  = 3.53 A, i = 3.53 sin(2π60t + 20)
b.
P = I2R =
c.
T =
a.
Im =
b.
L=
c.
L0W
a.
Em = ImXC = (30  103 A)(2.4 k) = 72 V
e = 72 sin(2π500t  20  90) = 72 sin(2π500t  110)
b.
C=
c.
P=0W
33.
34.
35.
a.
2
3.53 A
2
2
6.8 Ω = 42.38 W
6.28
= 16.67 ms
ω 2 60 rad/s
6(16.67 ms) = 100.02 ms  0.1 s
=
Vm 128 V
= 4.27 A, i = 4.27 sin(1000t  30)
=
X L 30 
XL
ω
=
30 
= 30 mH, standard = 30 mH
1000 rad/s
1
1
= 0.133 μF standatd = 0.13 μF
=
3
ωX C (3.14  10 rad/s)(2.4 k)
1
1
1
= 50 
=
=
4
2 f C1 ωC1 (10 rad/s)(2  F)
1
1
= 10 
X C2 =
=
4
ωC 2 (10 )(10  F )
E
84.85 V 60
=
= 1.697 A 150
E = 84.85 V 60
I1 =
50    90
Z C1
X C1 =
I2 =
E
ZC 2
=
84.85 V 60
= 8.485 A 150
10    90
i1 = 2.4 sin(10 t + 150)
i2 = 12 sin(104t + 150)
4
CHAPTER 14
169
b.
CT  2  F  10  F  12  F
1
1
XC 

4
 C (10 rad/s)(12  F)
 8.33 
E
84.8560
Is =

XCT 8.33   90
= 10.19 A150
is = 14.4 sin (104t + 150°)
36.
a.
L1  L2 = 60 mH  120 mH = 40 mH
3
X LT = 2πfLT = 2π(10 Hz)(40 mH) = 251.33 
Vm = I m X LT =  24 A  (251.33 ) = 6.03 kV
and υs = 6.03 kV sin(103t + 30 + 90)
or υs = 6.03  103 sin(103t + 120)
b.
I m1 =
Vm
, X L1 = 2fL1 = 2(103 Hz)(60 mH) = 376.99 
X L1
I m1 =
6.03  103 V
= 16 A
376.99 
and i1 = 16 sin(103t + 30)
3
X L2 = 2πfL2 = 2π(10 Hz)(120 mH) = 753.98 Ω
I m2 =
6.03  103 V
=8A
753.98 
and i2 = 8 A sin(103t + 30)
37.
a.
c.
e.
5.0 36.87
12.65 7.57
4123.11 104.04
b.
d.
f.
2.83 45
1001.25 2.86
0.894 116.57
38.
a.
c.
e.
17.89 116.57
20.22 × 103 8.53
 200 0°
b.
d.
f.
8.94 26.57
8.49 × 103 135
1000 178.85
39.
a.
c.
e.
4.6 + j3.86
j2000
47.97 + j1.68
b.
d.
f.
6.0 + j10.39
6 × 103  j2.2 × 103
4.7 × 104 j1.71 × 104
40.
a.
c.
e.
42 + j0.11
3 × 103  j5.20 × 103
15
b.
d.
f.
1 × 103  j1.73 × 103
6.13 × 103 + j5.14 × 103
2.09 × 103  j1.20
170
CHAPTER 14
41.
42.
43.
44.
45.
46.
47.
48.
a.
11.8 + j7.0
b.
151.90 + j49.90
c.
4.72 × 106 + j71
a.
5.20 + j1.60
b.
209.30 + j311.0
c.
21.20 + j12.0
a.
12.17 54.70°
b.
98.37 13.38°
c.
28.07 115.91°
a.
12.0 + j34.0
b.
86.80 + j312.40
c.
283.90  j637.65
a.
8.00 20°
b.
49.68 64.0°
c.
40 × 10340°
a.
6.0 50°
b.
200 × 106 60°
c.
109 170°
a.
4
b.
4.15  j4.23
c.
6.69  j6.46
a.
10  j 5
= 10.0  j5.0
1  j0
b.
8 60
8 60
= 19.38  103 15.69

102  j 400 412.80 75.69
c.
(6 20)(120   40)(8.54 69.44) 6.15  103 49.44
= 3.07  103 79.44

2   30
2   30
CHAPTER 14
171
49
a.
(0.16 120)(300 40)
48 160

= 5.06 88.44
9.487 71.565
9.487 71.565
b.

1
1

 8 



4
2 
 4  10 20   j ( j )   36  j 30 
 8 
1







46.861
39.81
j


 
 2500   20  
(2500 20)(8j)(0.0213 39.81) = 426 109.81
50.
51.
a.
x + j4 + 3x + jy  j7 = 16
(x + 3x) + j(4 + y  7) = 16 + j0
x + 3x = 16
4+y7=0
4x = 16
y = +7  4
x=4
y=3
b.
(10 20)(x 60) = 30.64  j25.72
10x 40 = 40 40
10 x = 40
x=4
5x + j10
2  jy
──────
10x + j20  j5xy  j210y = 90  j70
(10x + 10y) + j(20  5xy) = 90  j70
10x + 10y = 90
x+y=9
x=9y
a.
20  5xy = 70
20  5(9  y)y = 70
5y(9  y) = 90
y2  9y + 18 = 0
(9)  (9) 2  4(1)(18)
2
93
y=
= 6, 3
2
y=
For y = 6, x = 3
y = 3, x = 6
(x = 3, y = 6) or (x = 6, y = 3)
b.
52.
172
80 0
= 4 θ = 3.464  j2 = 4 30
40 
θ = 30
a.
160.0 30
b.
25  103 40
c.
70.71 90
CHAPTER 14
53.
54.
a.
14.14 180
b.
4.24  106 90
c.
2.55 × 10670
a.
56.57 sin(377t + 20)
b.
169.68 sin (377t + 10)
c.
11.31  103 sin(377t  110)
d.
6000 sin(377t  180)
55.
(Using peak values)
ein = υa + υb  υa = ein  υb
= 60 V 45  20 V 45
= 63.25 V  63.43
and ein = 63.25 sin (377t + 63.43)
56.
is = i1 + i2  i1 = is  i2
(Using peak values) = (20  106 A 60)  (6  106 A 30) = 20.88  106 A  76.70
i1 = 20.88  106 sin (t + 76.70)
57.
ein = υa + υb + υc
υa = ein  υb  υc
= 120 V 30°  30 V 60°  40 V 120°
= 108.92 V 0.33°
ein = 108.92 sin(377t  0.33°)
58.
Is = I1 + I2 + I3
I1 = Is  I2  I3
= 12.73 A 180°  5.66 A 180°  2[5.66 A 180°]
= 12.73 A 180°  5.66 A 180°  11.32 A 180°
= 4.25 A 0°
i1 = 6.01 sin 377t
CHAPTER 14
173
Chapter 15
1.
2.
a.
R  0 = 6.8   0 = 6.8 
b.
XL = L = (377 rads/s)(1.2 H) = 452.4 
XL  90 = 452.4   90 = +j452.4 
c.
XL = 2fL = (6.28)(50 Hz)(47 mH) = 1.48 
XL  90 = 1.48   90 = +j1.48 
d.
XC =
C (100 rad/s)(10  106 F)
XC  90 = 1 k  90 = j1 k
e.
XC =
f.
R  0 = 220   0 = 220 
a.
V = 10.61 V  10, I =
1

1
1
1
= 33.86 

3
2 fC 2 (10  10 Hz)(0.47  F)
XC  90 = 33.86   90 = j33.86 
V  10.61 V 10
= 3.54 A  10

R0
3  0
i = 5 sin (t + 10)
b.
= 1 k
V = 4.24 V  10, I =
V 
4.24 V 10
= 4.24 A  80

X L 90
1  90
i = 6 sin (t  80)
c.
3.
1
1
= 15.924 

2 fC 2 (5 kHz)(2  F)
V 
84.84 V 0
= 5.328 A 90
I=

X C   90 15.924    90
i = 7.534 sin (t + 90)
V = 84.84 V  0, XC =
a.
I = (0.707)(4 mA  0) = 2.828 mA  0
V = (I  0)(R  0) = 2.828 mA  0)(22   0) = 62.216 mV  0
 = 88  103 sin 1000t
b.
I = (0.707)(1.5 A  60) = 1.061 A  60
XL = 2πfL = 2π(200 Hz)(12 mH) = 15.08 Ω
V = (I  )(XL  90) = (1.061 A  60)(15.08   90) = 16 V  150
 = 22.62 sin(2π200t + 150)
174
CHAPTER 15
c.
I = (0.707)(2 mA  40) = 1.414 mA  40
1
1

= 135.52 k
XC =
 C (157rad/s)(0.047  F)
V = (I  )(XC  90) = (1.414 mA  40)(135.52 k  90) = 191.63 V  50
Vp = 2(191.63 V) = 270.96 V
and  = 270.96 sin (157t  50)
4.
5.
6.
7.
a.
ZT = 6.8  + j8.2  = 10.65   50.33
b.
ZT = 2   j6  + 10  = 12   j6  = 13.42   26.57
c.
ZT = 1 k + j3.2 k + 5.6 k + j6.8 k = 6.6 k + j10 k = 11.98 k  56.58
a.
ZT = 3  + j4   j5  = 3   j1  = 3.16   18.43
b.
ZT = 1 k + j8 k  j4 k = 1 k + j4 k = 4.12 k  75.96
c.
LT = 247 mH
XL = L = 2fL = 2(103 Hz)(247  103 H) = 1.55 k
1
1
= 1.59 k
XC =

3
2 fC 2 (10 Hz)(0.1  106 F)
= 470  + j1.55 k  j1.59 k
= 470   j40  = 471.70   4.86
E 120 V 0
= 2   70 = 0.684   j1.879  = R  jXC

I 60 A 70
a.
ZT =
b.
ZT =
c.
ZT =
a.
ZT = 8  + j6  = 10   36.87
c.
I = E/ZT = 100 V  0/10   36.87 = 10 A  36.87
VR = (I  )(R  0) = (10 A  36.87)(8   0) = 80 V  36.87
VL = (I  )(XL  90) = (10 A  36.87)(6   90) = 60 V  53.13
f.
P = I2R = (10 A)2 8  = 800 W
g.
Fp = cos θT = R/ZT = 8 /10  = 0.8 lagging
h.
R = 113.12 sin(t  36.87)
L = 84.84 sin(t + 53.13)
i = 14.14 sin (t  36.87)
CHAPTER 15
E 80 V 320
= 4 k  280 = 4 k  80 = 0.695 k  j3.939 

I 20 mA 40
= R  jXC
8 kV 0
E
= 40 k  60 = 20 k + j34.64 k = R + jXL

I 0.2 A   60
175
8.
a.
ZT = 18   j29.15  = 34.26  58.30
1
1

= 29.15 Ω
XC =
2 fC 2 (60 Hz)(91  F)
c.
I=
E
120 V  20
= 3.50 A 78.30
=
ZT 34.26    58.30
VR = (I θ)(R 0) = (3.50 A 78.30)(18  0) = 63.0 V 78.30
VC = (I θ)(XC 90) = (3.50 A 78.30)(29.15  90) = 102.03 V 11.70
9.
f.
P = I2R = (3.50 A)2 18  = 220.5 W
g.
Fp = R/ZT = 18 /34.26 Ω = 0.525 leading
h.
i = 4.95 sin(377t + 78.30)
υR = 89.1 sin(377t + 78.30)
υC = 144.27 sin(377t  11.70)
a.
ZT = 4  + j6   j10  = 4   j4  = 5.66  45
c.
XL = ωL  L =
d.
=
E
50 V 0
= 8.83 A 45
=
5.66    45
ZT
VR = (I θ)(R 0) = (8.83 A 45)(4  0) = 35.32 V 45
VL = (I θ)(XL 90) = (8.83 A 45)(6  90) = 52.98 V 135
VC = (I θ)(XC 90) = (8.83 A 45)(10  90) = 88.30 V 45
I=
f.
E = VR + VL + VC
50 V 0 = 35.32 V 45 + 52.98 V 135 + 88.30 V 45
50 V 0 = 49.95 V 0  50 V  0
g.
P = I2R = (8.83 A)2 4  = 311.88 W
h.
Fp = cos θT =
i.
176
6
= 16 mH
 377 rad/s
1
1
1
 C=
= 265 μF
XC =
=
C
 X C (377 rad/s)(10 )
XL
R
ZT
= 4 Ω/5.66 Ω = 0.707 leading
i = 12.49 sin(377t + 45)
e = 70.7 sin 377t
υR = 49.94 sin(377t + 45)
υL = 74.91 sin(377t + 135)
υC = 124.86 sin(377t  45)
CHAPTER 15
10.
11.
a.
XL = ωL = (20 × 103 rad/s)(0.1H) = 2 kΩ
1
1
= 6.1 kΩ
XC =

3
 C (20  10 rad/s)(8200 pF)
ZT = 1.2 kΩ + j2 kΩ  j6.1 kΩ
= 1.2 kΩ  j4.1 kΩ = 4.27 kΩ 73.69°
b.

c.

d.
I=
e.

f.
E = VR + VL + VC
4.24 V 60° = 1.19 V 133.69° + 1.99 V 223.69° + 6.06 V 43.69°
= (0.822 V + j0.80 V) + (1.44 V  j1.37 V) + (4.38 V + j 4.19 V)
= 2.12 V+ j3.62 V
4.24 V 60°  4.20 V 59.65°
E
4.24 V60
= 0.993 mA 133.69°

ZT 4.27 k  73.69
VR = IR = (0.993 mA 133.69°)(1.2 kΩ 0°) = 1.19 V 133.69°
VL = IXL = (0.993 mA 133.69°)(2 kΩ 90°) = 1.99 V 223.69°
VC = IXC = (0.993 mA 133.69°)(6.1 kΩ 90°) = 6.06 V 43.69°
g.
P = I2R = (0.993 mA)2(1.2 kΩ) = 1.18 mW
h.
Fp =
i.
i = 1.4 × 103 sin (20,000t + 133.69°)
υR = 1.68 sin (20,000t + 133.69°)
υL = 2.81 sin (20,000t + 223.69°)
υC = 8.57 sin (20,000t + 43.69°)
R
1.2 k
= 0.281 leading

ZT 4.27 k
20 V (rms)  28.28 V (peak)
43.20 V(p  p)  21.60 V (peak)
22 (28.28 V)
22   R
475.20 + 21.60R = 622.16
146.96 
= 6.8 Ω
R=
21.60
Vscope = 21.60 V =
CHAPTER 15
177
12.
a.
 22.8 V 
VL (rms) = 0.7071 
= 8.06 V
 2 
V (rms) 8.06 V
XL = L
= 6.2 kΩ

I (rms) 1.3 mA
XL = L = (1000rad/s)L = 6.2 kΩ  L =
b.
6.2 k
= 6.2 H
1000 rad/s
E 2  VR2  VL2
(22 V)2 = VR2  (8.06 V) 2
484  VR2  64.96
VR2  419.04
VR  419.04  20.47 V
R
c.
13.
a.
VR (rms) 20.47 V

 15.75 k
I (rms)
1.3 mA
6.2 H
 8.27 V 
= 2.924 V
VR(rms) = 0.7071 
 2 
V (rms) 2.924 V
= 292.4 μA
I(rms) = R

10 k
R2
E 2  VR2  VC2
b.
(12 V) 2  (2.924 V) 2  VC2
144  8.55  VC2
VC2  135.35
VC  135.45  11.64 V
VC (rms) 11.64 V

= 39.81 kΩ
I (rms) 292.4  A
1
1
1
= 100 pF
XC =
C 

2 fC
2 fX C 2 (40 kHz)(39.81 k)
XC =
14.
a.
b.
178
(2 k  0)(120 V 60) 240 V 60
= 29.09 V 15.96
=
2 k  + j8 k 
8.25 75.96
(8 k  90)(120 V 60)
= 116.36 V 74.04
V2 =
8.25 k  75.96
V1 =
( 40  90)(60 V 5) 2400 V 95
= 48.69 V 40.75
=
6.8  + j 40  + 22 
28.8 + j 40
(22  0)(60 V 5)
1.32 kV 5
= 26.78 V 49.25
V2 =
=
49.29  54.25
49.29  54.25
V1 =
CHAPTER 15
15.
a.
b.
16.
a.
ZT = 4.7 k + j30 k + 3.3 k  j10 k = 8 k + j20 k = 21.541 k 68.199
ZT = 3.3 k + j30 k  j10 k = 3.3 k + j20 k = 20.27 k 80.631
Z E (20.27 k 80.631)(120 V 0)
= 112.92 V 12.432
V1 = T =
ZT
21.541 k 68.199
Z E
ZT = 3.3 k  j10 k = 10.53 k 71.737
V2 = T
ZT
(10.53 k   71.737)(120 V 0)
= 58.66 V 139.94
=
21.541 k 68.199
XL = ωL = (1000 rad/s)(20 mH) = 20 
1
1
=
= 25.64 
XC =
ωC (1000 rad/s)(39  F)
ZT = 30  + j20   j25.64  = 30   j5.64  = 30.53  10.65
E
20 V  40
= 655.1 mA 50.65
I=
=
ZT 30.53    10.65
VR = (I  θ)(R  0) = (655.1 mA 50.65)(30  0) = 19.65 V 50.65
VC = (655.1 mA 50.65)(25.64  90) = 16.80 V 39.35
R
30 
= 0.983 leading
=
ZT 30.53 
b.
cos θT =
c.
P = I2R = (655.1 mA)2 30  = 12.87 W
f.
g.
17.
(20  90)(20 V 70)
= 14.14 V 155
20  + j 20   j 40 
(40    90)(20 V 70)
= 28.29 V 25
V2 =
28.28    45
V1 =
(30  0)(20 V  40)
600 V 40
= 19.66 V 50.65
=
ZT
30.53    10.65
(25.64    90)(20 V 40)
= 16.80 V 39.35
VC =
30.53    10.65
VR =
ZT = 30   j5.64  = R  jXC
P = VI cos θ  8000 W = (200 V)(I)(0.8)
8000 A
= 50 A
I=
160
0.8 = cos θ
θ = 36.87
V = 200 V 0, I = 50 A 36.87
V
200 V 0
= 4  36.87 = 3.2  + j2.4 
=
ZT =
I 50 A   36.87
CHAPTER 15
179
18.
P = VI cos   300 W = (120 V)(3 A) cos θ
cos θ = 0.833  θ = 33.59
V = 120 V 0, I = 3 A 33.59
V
120 V 0
=
= 40  33.59 = 33.34  + j22.10 
ZT =
I 3 A   33.59
RT = 33.34  = 2  + R  R = 31.34 
19.
a.
b.
ZT =
R 2 + X L2 tan1XL/R
ZT
1.0 k
1.008 k
1.181 k
1.606 k
2.134 k
2.705 k
XLE
ZT
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
VL
0.0 V
0.623 V
2.66 V
3.888 V
4.416 V
4.646 V
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
θL = 90  tan1 XL/R
90.0
82.84
57.85
38.5
27.96
21.7
VL =
c.
180
θT
0.0
7.16
32.14
51.49
62.05
68.3
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
CHAPTER 15
d.
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
20.
a.
ZT =
│ZT│ =
1
R  X C tan XC/R
2
2
R 2  X C2 , θT = tan1XC/R
f
0 kHz
1 kHz
3 kHz
5 kHz
7 kHz
10 kHz
CHAPTER 15
VR = RE/ZT
5.0 V
4.96 V
4.23 V
3.11 V
2.34 V
1.848 V
│ZT│

353.1 
150.80 
120.78 
111.09 
105.58 
θT
90.0
73.55
48.46°
34.11
25.82
18.71
181
b.
VC =
( X C   90)(E 0)
XC E
90 + tan1XC/R
=
2
2
R  jX C
R  XC
│VC│ =
XC E
R 2  X C2
│VC│
10.0 V
9.59 V
7.49 V
5.61 V
4.36 V
3.21 V
f
0 Hz
1 kHz
3 kHz
5 kHz
7 kHz
10 kHz
c.
θC = 90 + tan1 XC/R
θC
0.0
16.45
41.54
55.89
64.18
71.29
f
0 Hz
1 kHz
3 kHz
5 kHz
7 kHz
10 kHz
d.
│VR│ =
RE
R  X C2
f
0 Hz
1 kHz
3 kHz
5 kHz
7 kHz
10 kHz
182
2
│VR│
0.0 V
2.83 V
6.63 V
8.28 V
9.00 V
9.47 V
CHAPTER 15
21.
a.
R 2  ( X L  X C ) 2  tan 1 ( X L  X C ) / R
ZT =
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
b.
ZT

19.31 × 103 
3.40 × 103 
1.21 × 103 
1.16 × 103 
1.84 × 103 
│VC│ =
XC E
ZT
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
c.
CHAPTER 15
θT
90.0
87.03
72.91
34.33
+30.75
+56.99
E
ZT
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
│VC│
120 V
120.62 V
136.94 V
192.4 V
133.45 V
63.29 V
I =
I
0.0 mA
6.21 mA
35.29 mA
99.17 mA
103.45 mA
65.22 mA
183
22.
1
1
1
=Rf=
= 1.54 kHz

2 fC
2 RC 2 (220 )(0.47  F)
a.
XC =
b.
Low frequency: XC very large resulting in large ZT
High frequency: XC approaches zero ohms and ZT approaches R
c.
f = 100 Hz: XC =
1
1
= 3.39 k

2 fC 2 (100 Hz)(0.47  F)
ZT  XC
f = 10 kHz: XC =
1
1
= 33.86 

2 fC 2 (10 kHz)(0.47  F)
ZT  R
23.
d.

e.
f = 40 kHz: XC =
a.
YT =
b.
c.
24
184
a.
1
1
= 8.47 k

2 fC 2 (40 kHz)(0.47  F)
X
8.47 
 = tan1 C   tan 1
= 2.2
R
220 
1
1
 0 
0 = 0.147 S 0°
R
6.8 
1
1
YT =

  90 = 5 mS 90°
X L 90 200 
YT =
1
1
1


90 = 0.5 mS 90°
ZT X C   90 2 k
(10 0)(60  90)
= 9.86  9.46
10   j 60 
1
1
YT =
= 0.10 S 9.46° = 0.1 S  0.02 S = G  jBL

ZT 9.86 9.46
ZT =
b.
22   2.2  = 2 
(2 0)(6    90)
12    90
= 1.90  18.43
ZT =

2   j6 
6.32    71.57
1
1
YT =
= 0.53 S 18.43° = 0.5 S + j0.17 S = G + jBC

ZT 1.90   18.43
c.
YT =
1
1
1
+
+
3 k  0 6 k  90 9 k    90
= 0.333  103 0 + 0.167  103 90 + 0.111  103 90
= 0.333  103 S  j0.056  103 S = 0.34 mS 9.55
= G  jBL
CHAPTER 15
25.
a.
ZT = 4.7  + j8  = 9.28   59.57, YT = 0.108 S 59.57
YT = 54.7 mS  j93.12 mS = G  jBL
b.
ZT = 33  + 20   j70  = 53   j70  = 87.80  52.87
YT = 11.39 mS 52.87 = 6.88 mS + j9.08 mS = G + jBC
c.
ZT = 200  + j500   j600  = 200   j100  = 223.61  26.57
YT = 4.47 mS 26.57 = 4 mS + j2 mS = G + jBC
26.
a.
YT =
b.
YT =
I 60 A 70
=
= 0.5 S 70 = 0.171 S + j0.470 S = G + jBC
E 120 V 0
1
1
= 5.85 , XC =
= 2.13 
R=
G
BC
I 20 mA 40
=
= 0.25 mS 280 = 0.25 mS 80
E 80 V 320
= 0.043 mS + j0.246 mS = G + jBC
1
1
= 23.26 k, XC =
= 4.07 k
R=
G
BC
27.
I 0.2 A   60
=
= 0.25 mS 60 = 0.0125 mS  j0.02165 mS = G  jBL
E
8 kV 0
1
1
R=
= 80 k, XL =
= 46.19 k
G
BL
c.
YT =
a.
YT =
c.
E = Is/YT = 2 A 0/111.8 mS 26.57= 17.89 V 26.57
E 
= 17.89 V 26.57/10  0 = 1.79 A 26.57
IR =
R 0
E 
IL =
= 17.89 V 26.57/20  90 = 0.89 A 63.43
X L 90
f.
P = I2R = (1.79 A)2 10  = 32.04 W
g.
Fp =
h.
1
1
+
= 0.1 S  j0.05 S = 111.8 mS 26.57
10  0 20  90
G
YT
=
0.1 S
= 0.894 lagging
111.8 mS
e = 25.30 sin(377t + 26.57)
iR = 2.53 sin(377t + 26.57)
iL = 1.26 sin(377t  63.43)
is = 2.83 sin 377t
CHAPTER 15
185
28.
a.
XC =
b.
YT =
c.
29.
1
1

= 20.4 kΩ
2 fC 2 (60 Hz)(0.13  F)
1
1
+
= 0.1 mS 0 + 0.049 mS 90
10 k  0 20.4 k    90
= 0.111 mS 26.10
2 mA 20
Is
= 18.02 V 6.1
=
0.111 mS 26.10
YT
E
18.02 V   6.1
=
= 1.80 mA 6.1
IR =
10 k  0
ZR
E
18.02 V   6.1
IC =
=
= 0.883 mA 83.90
20.4 k    90
ZC
E=
e.
Is = IR + IC
2 mA 20 = 1.80 mA 6.1 + 0.883 mA 83.90
= (1.79 mA  j0.191 mA) + (0.094 mA + j0.878 mA)
= 1.88 mA + j0.687 mA
2 mA 20°  2 mA 20.07°
f.
P = I2R = (1.80 mA)2 10 k = 32.4 mW
g.
Fp =
h.
ω = 2f = 377 rad/s
a.
c.
G
YT
=
0.1 mS
= 0.9 leading
0.111 mS
is = 2.83  103 sin(ωt + 20)
iR = 2.55  103 sin(ωt  6.57)
iC = 1.25  103 sin(ωt + 83.44)
e = 25.48 sin(ωt  6.57)
1
1
1


1.2  0 2  90 5    90
= 0.833 S 0 + 0.5 S 90 + 0.2 S 90
= 0.833 S  j0.3 S = 0.89 S 19.81
ZT = 1.12   19.81
YT =
XC =
1
1
1
 C=
= 531 μF
=
C
 X C (377 rad/s)(5 )
XL = ωL  L = X L =
ω
186
2
= 5.31 mH
377 rad/s
CHAPTER 15
d.
g.
Is = IR + IL + IC
2.121 A 60 = 2.00 A 79.81 + 1.20 A 10.19 + 0.48 A 169.81

2.121 A 60 = 2.13 A 60.01
2
P = I R = (2.00 A)2 1.2  = 4.8 W
h.
Fp =
f.
i.
30.
(0.707)(3 A) 60
2.121 A 60
=
E = Is =
= 2.40 V 79.81
0.885 S   19.81 0.885 S   19.81
YT
E  2.397 V 79.81
IR =
= 2.00 A 79.81
=
R 0
1.2  0
E 
2.397 V 79.81
IL =
=
= 1.20 A 10.19
2  90
X L 90
E 
2.397 V 79.81
IC =
=
= 0.48 A 169.81
5    90
X C   90
a.
G
YT
=
0.833 S
= 0.941 lagging
0.885 S
e = 3.39 sin(377t + 79.81)
iR = 2.83 sin(377t + 79.81)
iL = 1.70 sin(377t  10.19)
iC = 0.68 sin(377t + 169.81)
XL = L = (1000 rad/s)(3.9 H) = 3.9 kΩ,
1
1

= 8.33 kΩ
XC =
 C (1000 rad/s)(0.12  F)
1
1
1
YT =
+
+
3 k  0 3.9 k  90 8.33 k    90
= 0.333 mS 0 + 0.256 mS 90 + 0.120 mS 90
= 0.333 mS  j0.136 mS = 0.36 mS 22.22
d.
E = I/YT = 3.54 mA 20/0.36 mS 22.22 = 9.83 V 2.22
E 
IR =
= 9.83 V2.22/3 k 0 = 3.28 mA 2.22
R 0
E 
IL =
= 9.83 V2.22/3.9 k 90 = 2.52 mA 87.78
X L 90
E 
IC =
= 9.83 V2.22/8.33 k 90 = 1.18 mA 92.22
X C   90
g.
P = I2R = (3.28 mA)23 k = 32.28 mW
h.
Fp = G/YT = 0.333 mS/0.36 mS = 0.925 leading
i.
e = 13.9 sin(1000t + 2.22)
iR  4.64  103 sin(1000t + 2.22)
iL  3.56  103 sin(1000t  87.78)
iC = 1.67  103 sin(1000t + 92.22)

CHAPTER 15
187
31.
a.
b.
(60  90)(20 A 40) 1200 A 130
= 18.78 60.14°

22   j 60 
63.91 69.86
(22  0)(20 A 40)
440 A 40

I2 =
= 6.88 29.86°
63.91  69.86
63.91 69.86
I1 =
(12   j 6 )(6 A 30) (13.42   26.57)(6 A 30)

12   j 6   j 4 
12  j 2
80.52 A 3.43
= 6.62 A 12.89°
=
12.17   9.46
I1 =
I2 =
c.
32.
a.
(4  90)(6 A 30)
24 A 120
= 1.97 A 129.46°

12.17   9.46
12.17   9.46
( j10   j 40 )(4 A 0) (30  90)(4 A 0)

 j 20   j10   j 40 
50  90
= 2.4 A 0°
I1 =
I2 =
(20  90)(4A 0)
= 1.6 A 0°
50  90
ZT =
(R 0 )( X C   90 )
= RX C 90 + tan1XC/R
2
2
R  jX C
R  XC
│ZT│ =
RX C
R
f
0 Hz
1 kHz
2 kHz
3 kHz
4 kHz
5 kHz
10 kHz
20 kHz
188
2
 X C2
│ZT│
40.0 
35.74 
28.22 
22.11 
17.82 
14.79 
7.81 
3.959 
θT = 90 + tan1XC/R
θT
0.0
26.67
45.14
56.44
63.55
68.30
78.75
89.86
CHAPTER 15
b.
IRX C
│VC│ =
R 2 + X C2
│VC│
2.0 V
1.787 V
1.411 V
1.105 V
0.891 V
0.740 V
0.391 V
0.198 V
f
0 kHz
1 kHz
2 kHz
3 kHz
4 kHz
5 kHz
10 kHz
20 kHz
c.
│IR│ =
VC
R
│IR│
50.0 mA
44.7 mA
35.3 mA
27.64 mA
22.28 mA
18.50 mA
9.78 mA
4.95 mA
f
0 kHz
1 kHz
2 kHz
3 kHz
4 kHz
5 kHz
10 kHz
20 kHz
33.
a.
ZR ZL
( R 0)( X L 90)
=
=
ZR  ZL
R  jX L
ZT =
│ZT│ =
RX L
R
CHAPTER 15
= I[ZT(f)]
2
 X L2
RX L
R
2
 X L2
90  tan1XL/R
θT = 90  tan1XL/R
f
| ZT |
θT
0 Hz
1 kHz
5 kHz
7 kHz
10 kHz
0.0 k
1.22 k
3.91 k
4.35 k
4.65 k
90.0
75.86
38.53
29.6
21.69
189
│IL│ =
b.
34.
f
| IL |
0 Hz
1 kHz
5 kHz
7 kHz
10 kHz

31.75 mA
6.37 mA
4.55 mA
3.18 mA
IR =
c.
YT =
E
XL
E 40 V
= 8 mA (constant)
=
R 5k
R 2  X C2
RX C
90  tan1XC/R
f
│YT│
0 Hz 25.0 mS
1 kHz 27.98 mS
2 kHz 35.44 mS
3 kHz 45.23 mS
4 kHz 56.12 mS
5 kHz 67.61 mS
10 kHz 128.04 mS
20 kHz 252.59 mS
35.
YT =
0.0
26.67
45.14
56.44
63.55
68.30
78.75
89.86
1
(use data of Prob. 36),  TY =  TZ
ZT
f
0 Hz
1 kHz
5 kHz
7 kHz
10 kHz
190
θT
YT

0.82 mS
0.256 mS
0.23 mS
0.215 mS
θT
90.0
75.86
38.53
29.6
21.69
CHAPTER 15
36.
a.
YT = G 0 + BL 90 + BC 90
B  BL
= G 2  ( BC  BL ) 2 tan1 C
G
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
│YT│
XL  0 , ZT = 0 ,
YT =  
1.86 mS
1.02 mS
1.00 mS
1.02 mS
1.04 mS
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
│θT│
90.0
57.51
12.63
+1.66
+9.98
+16.54
f
0 Hz
b.
ZT =
1
YT
f
0 kHz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
CHAPTER 15
, T Z =   T Y
ZT
0.0 
537.63 
980.39 
1 k
980.39 
961.54 
θT
90.0
57.52
12.63
1.66
9.98
16.54
191
c.
VC(f) = I[ZT(f)]
│VC│
0.0 V
5.38 V
9.80 V
10 V
9.80 V
9.62 V
f
0 kHz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
d.
IL =
VL (f) VC ( f )

XL
XL
f
0 kHz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
37.
a.
b.
IL
10.0 mA
8.56 mA
3.12 mA
1.59 mA
1.04 mA
0.765 mA
Rp =
Rs2  X s2 (20 ) 2  (40 ) 2
= 100  (R)

20 
Rs
Xp =
Rs2  X s2 2000 
= 50  (C)

40
Xs
Rp =
Rs2  X s2 (2 k ) 2 + (8 k ) 2
= 34 k (R)
=
Rs
2 k
Rs2  X s2 (2 k ) 2 + (8 k ) 2
= 8.5 k (L)
Xp =
=
8k
Xs
38.
a.
R p X p2
(8.2 k )(20 k ) 2
=
= 7.02 k
Rs = 2
X p  R p2 (20 k ) 2 + (8.2 k ) 2
Xs =
R p2 X p
=
2
X p2  R p
(8.2 k ) 2(20 k )
= 2.88 k
467.24 k 
ZT = 7.02 k  j2.88 k
b.
Rs =
Xs =
R p X p2
=
2
X p2  R p
R p2 X p
(68 )(40 ) 2
= 17.48 
(40 ) 2 + (68 ) 2
=
2
X 2p  R p
(68 ) 2(40 )
= 29.72 
6224  2
ZT = 17.48  + j29.72 
192
CHAPTER 15
39.
a.
CT = 2 μF
1
1
= 79.62 
XC =
=
3
C 2 (10 Hz)(2  F)
XL = ωL = 2π(103 Hz)(10 mH) = 62.80 
YT =
1
1
1


220  0 79.62    90 62.8  90
= 4.55 mS 0 + 12.56 mS 90 + 15.92 mS 90
= 4.55 mS  j3.36 mS = 5.66 mS 36.44
E = I/YT = 1 A 0/5.66 mS 36.44 = 176.68 V 36.44
E 
= 176.68 V 36.44/220  0 = 0.803 A 36.44
IR =
R 0
E 
= 176.68 V 36.44/62.80 90 = 2.813 A 53.56
IL =
X L 90
40.
b.
Fp = G/YT = 4.55 mS/5.66 mS = 0.804 lagging
c.
P = I2R = (0.803 A)2 220  = 141.86 W
f.
Is = IR + 2IC + IL
I  I  IL
and IC = s R
2
1 A 0  0.803 A 36.44  2.813 A   53.56
=
2
1  (0.646  j 0.477)  (1.671  j 2.263) 1.317  j1.786
=

2
2
IC = 0.657 + j0.893 = 1.11 A 126.43
g.
ZT =
1
1
= 176.7  36.44
=
YT
5.66 mS   36.44
= 142.15  + j104.96  = R + jXL
P = VI cos θ = 3000 W
3000 W
3000 W
3000
=
=
= 0.75 (lagging)
cos θ =
(100 V)(40 A) 4000
VI
θ = cos1 0.75 = 41.41
I
40 A   41.41
=
= 0.4 S 41.41 = 0.3 S  j0.265 S = GT  jBL
YT =
E
100 V 0
1
1
1
GT = 0.3 S =
+ = 0.05 S +
20  R
R
1
and R =
=4
0.25 S
CHAPTER 15
193
XL =
41.
1
BL

1
= 3.74 
0.265 S
a.
b.
c.
42.
194
a.
e and  R2
b.
e and is
CHAPTER 15
43.
c.
iL and iC
(I):
(a)
θdiv. = 0.8 div., θT = 4 div.
0.8 div.
 360 = 72
θ=
4 div.
1 leads 2 by 72
(b)
1: peak-to-peak = (5 div.)(0.5 V/div.) = 2.5 V
 2.5 V 
V1(rms) = 0.7071 
 = 0.88 V
 2 
2: peak-to-peak = (2.4 div.)(0.5 V/div.) = 1.2 V
 1.2 V 
V2(rms) = 0.7071 
 = 0.42 V
 2 
(II):
(c)
T = (4 div.)(0.2 ms/div.) = 0.8 ms
1
1
=
= 1.25 kHz (both)
f=
T
0.8 ms
(a)
θdiv. = 2.2 div., θT = 6 div.
2.2 div.
 360 = 132
θ=
6 div.
1 leads 2 by 132
(b)
1: peak-to-peak = (2.8 div.)(2 V/div.) = 5.6 V
 5.6 V 
V1(rms) = 0.7071 
 = 1.98 V
 2 
2: peak-to-peak = (4 div.)(2 V/div.) = 8 V
8V
V2(rms) = 0.7071 
 = 2.83 V
 2 
(c)
CHAPTER 15
T = (6 div.)(10 s/div.) = 60 μs
1
1
=
= 16.67 kHz
f=
T
60  s
195
Chapter 16
1.
2.
ZT = j4  +
b.
Is =
c.
I1 = 3.5 A 22.65
d.
I2 =
e.
VL = Is XL = (3.5 A 22.65)(4  90) = 14 V 112.65
a.
ZT = 3  + j6  + 2  0  8  90
= 3  + j6  + 1.94  14.04
= 3  + j6  + 1.88   j0.47 
= 4.88  + j5.53  = 7.38  48.57
b.
Is =
c.
IC =
=
d.
3.
a.
b.
c.
196
(8    90)(12   0)
= 3.69   j1.54  4 22.65
 j8  12
a.
E
14 V 0
= 3.5 A 22.65
=
4    22.65
ZT
(8    90)(3.5 A 22.65)
= 1.94 A 33.66
12   j8 
E
30 V 0
= 4.07 A 48.57
=
ZT 7.38  48.57
Z R2 I s
Z R2 + Z C
=
(2  0)(4.07 A   48.57)
2   j8 
8.14 A   48.57
= 0.987 A 27.39
8.25   75.96
Z L E (6  90)(30 V 0)
180 V 90
=
=
ZT
7.38  48.57
7.38  48.57
= 24.39 V 41.43
VL =
ZT = 12  90  (9.1   j12 Ω) = 12  90  15.06  52.826
180.72 37.17
=
9.100
= 19.86  37.17
Is =
E
ZT
=
60 V 0
= 3.02 A 37.17
19.86  37.17
(12 Ω 90°)(3.02 A   37.17) 36.24 A 52.83
=
j12  + 9.1   j12 
9.1 0
= 3.98 A 52.83
(CDR) I2 =
CHAPTER 16
4.
5.
(12    90)(60 V 0)
720 V   90
=
9.1   j12 
15.06   52.826
= 47.81 V 37.17
d.
(VDR) VC =
e.
P = EI cos θ = (60 V)(3.02 A)cos(37.17)
= 181.20(0.797) = 144.42 W
a.
 6.8 k

0   8 k 90 
(4 k   90)(6 k 90)  2


ZT = 2 k 0 +
6.8 k
 j 4 k  j 6 k
 j 8 k
2
24 k 0
27.2 k 90
= 2 k +

2 90
3.4 k  j8 k

8.69 k 66.97
= 2 k + j12 k 3.13
k
23.03




2.88 k  j1.22 k
= 2 k + 7.88 kΩ  j12 k + j1.22 k
ZT = 4.88 k  j10.78 k = 11.83 k 65.64
b.
V2 = IZT' = (4 mA 0)(3.13 k 23.03) = 12.52 V 23.03
V
12.52 V 23.03)
= 1.57 A 66.97
IL = 2 
X L2
8 k 90
c.
Fp =
a.
400 Ω 90  400  90 =
b.
VC =
c.
P = EI cos θ = (100 V)(0.25 A) cos 36.86
= (25)(0.8) = 20 W
R
4.88 k
= 0.413 (leading)

ZT 11.83 k
400    90
= 200  90
2
Z = 100   j200  = 223.61  63.43
Z = j200  + j600  = +j400  = 400  90
(223.61    63.43)(400  90) 89444.00 26.57
ZT = Z  Z =

(100   j 200 )  j 400 
223.61 63.43
= 400  36.86
E
100 V 0
= 0.25 A 36.86
=
I=
400    36.86
ZT
CHAPTER 16
(200    90)(100 V 0) 20, 000 V   90
= 89.44 V 26.57
=
100   j 200 
223.61   63.43
197
6.
a.
Z1 = 3  + j4  = 5  53.13
E
120 V 0
= 24 A 53.13
I1 =
=
Z1 5  53.13
b.
VC =
c.
VR1 = (I1 θ)R 0 = (24 A 53.13)(3  0) = 72 V 53.13
(13    90)(120 V 0) 1560 V   90
= 260 V 0
=
 j13  + j 7 
6   90
Vab + VR1  VC = 0
+-
Vab = VC  VR1 = 260 V 0  72 V 53.13
+-
= 260 V  (43.20 V + j57.60 V)
= 216.80 V + j57.60 V = 224.32 V 14.88
7.
Z1 = 10  0
Z2 = 80  90  20  0
1600  90
1600  90

=
20  j80
82.462 75.964
= 19.403  14.036
Z3 = 60  90
a.
ZT = (Z1 + Z2)  Z3
= (10  + 18.824  + j4.706 )  60  90
1752.36    80.727
= 29.206  9.273  6  90 =
28.824  j 4.706  j 60
1752.36    80.727
=
= 28.103  18.259
62.356   62.468
E
40 V 0
= 1.42 A 18.26
I1 =
=
ZT 28.103    18.259
198
Z 2E
(19.403  14.036)(40 V 0) 776.12 V 14.036
=
=
29.206  9.273
29.206 9.273
Z 2 + Z1
= 26.57 V 4.76
b.
V1 =
c.
P = EI cos θ = (40 V)(1.423 A)cos 18.259
= 54.07 W
CHAPTER 16
8.
a.
Z1 = 2  + j1  = 2.236  26.565, Z2 = 3  0
Z3 = 16  + j15   j7  = 16  + j8  = 17.889  26.565
1
1
1
1
1
1
YT =
+
+
=
+
+
Z1 Z 2 Z 3 2.236  26.565 3  0 17.889  26.565
= 0.447 S 26.565 + 0.333 S 0 + 0.056 S 26.565
= (0.4 S  j0.2 S) + (0.333 S) + (0.05 S  j0.025 S)
= 0.783 S  j0.225 S = 0.82 S 16.03
1
1
= 1.23  16.03
ZT =
=
YT
0.82 S   16.03
b
I1 =
E
60 V 0
= 26.83 A 26.57
=
Z1 2.236  26.565
E 60 V 0
= 20 A 0
=
Z 2 3  0
E
60 V 0
I3 =
= 3.35 A 26.57
=
Z3 17.889  26.565
I2 =
c.
Is =
E
60 V 0
= 48.9 A 16.03
=
ZT 1.227  16.032
?
Is = I1 + I2 + I3
?
48.9 A 16.03 = 26.83 A 26.57 + 20 A 0 + 3.35 A 26.57
= (24 A  j12 A) + (20 A) + (3 A  j1.5 A)

= 47 A + j13.5 A = 48.9 A 16.03 (checks)
d.
9.
a.
G 0.783 S
= 0.955 (lagging)
=
YT 0.820 S
X L1 = ωL1 = 2(103 Hz)(0.1 H) = 628 
Fp =
X L2 = ωL2 = 2(103 Hz)(0.2 H) = 1.256 k
1
1
= 0.159 k
=
3
C 2 (10 Hz)(1  F)
ZT = R 0 + X L1 90 + XC 90  X L2 90
XC =
= 300  + j628  + 0.159 k 90  1.256 k 90
= 300  + j628   j182 
= 300  + j446  = 537.51  56.07
b.
Is =
CHAPTER 16
E
50 V 0
= 93 mA 56.07
=
ZT 537.51  56.07
199
c.
(CDR):
I1 =
Z L2 I s
Z L2 + ZC
=
(1.256 k  90)(93 mA   56.07)
+ j1.256 k   j 0.159 k 
116.81 mA 33.93
= 106.48 mA 56.07
1.097 90
ZC I s
(0.159 k    90)(93 mA   56.07)
I2 =
=
Z L2 + Z C
1.097 k  90
=
14.79 mA   146.07
= 13.48 mA 236.07
1.097 90
= 13.48 mA 123.93
=
d.
V1 = (I2 θ)( X L2 90) = (13.48 mA 123.92)(1.256 k 90)
= 16.93 V 213.93
Vab = E  (Is θ)(R 0) = 50 V 0  (93 mA 56.07)(300  0)
+-
= 50 V  27.9 V 56.07
= 50 V  (15.573 V  j23.149 V)
= 34.43 V + j23.149 V = 41.49 V 33.92
10.
e.
P = I s2 R = (93 mA)2300  = 2.595 W
f.
Fp =
a.
ZT = 1.2 k +
R
300 Ω
= 0.558 (lagging)
=
ZT 537.51 Ω
= 1.2 k +
(1.2 k 0)(1.8 k   90) 2.4 k 90

1.2 k  j1.8 k
2
2.16 k  90
 1.2 k 90
2.16  56.31
= 1.2 k + 1 k 33.69 + j1.2 k
= 1.2 k + 832.05   j554.70  + j1.2 k
= 2.03 k + j645.30  = 2.13 k 17.63
200
b.
V1 = IR1 = (20 mA 0)(1.2 k 0) = 24 V0
c.
I1 =
d.
 (2.4 k)

V2 = I(X L1  X L2 ) = (20 mA 0) 
90  = 24 V 90


2
e.
Vs = IZT = (20 mA 0)(2.13 k 17.63) = 42.60 V 17.63
(1.2 k 0)(20 mA 0)
2.4 A 0

= 11.11 mA 56.31
1.2 k  j1.8 k
2.16  103   56.31
CHAPTER 16
11.
Z1 = 2   j2  = 2.828  45
Z2 = 3   j9  + j6 
= 3   j3  = 4.243  45
Z3 = 10  0
a.
1
1
1
1
1
1
+
+
=
+
+
Z1 Z 2 Z 3
2.828    45
4.243    45 10  0
= 0.354 S 45 + 0.236 S 45 + 0.1 S 0 = 0.59 S 45 + 0.1 S 0
= 0.417 S + j0.417 S + 0.1 S
YT = 0.517 S + j 0.417 S = 0.66 S 38.89
1
1
= 1.52  38.89
ZT =
=
YT
0.66 S 38.89
YT =
12.
b.
V1 =
c.
I1 =
d.
Is =
(2  0)(60 V0)
120 V 0
= 42.43 V 45

2   j 
2.828   45
60 V 0
60 V 0
60 V 0
E



Z  3   j 9   j 6  3   j 3  4.243   45
= 14.14 A 45
60 V 0
E
= 39.47 A 38.89

ZT 1.52    38.89
Z = 12   j20  = 23.32  59.04
R4 0  Z = 20  0  23.32  59.04 = 12.36  27.03
Z = R3 0 + R4 0  Z = 12  + 12.36  27.03
= 12  + (11.01   j5.62 )
= 23.01   j5.62  = 23.69  13.73
R2 0  Z = 20  0  23.69   13.73 = 10.92  6.29
ZT = R1 0 + R2 0  Z = 12  + 10.92 6.29
= 12  + (10.85   j1.25 )
= 22.85   j1.2  = 22.88  3.01
E
100 V 0
Is =
= 4.37 A  +3.01
=
ZT
22.88    3.01
I R1 = I
I R3 =
R2 0 I s
(20  0)(4.37 A 3.01) 87.40 A 3.01
=
=
43.38   7.44
R2 0  Z " 20   23.01   j 5.62 
= 2.01 A 10.45
R4 0I R3
(20  0)(2.01 A 10.45) 40.20 A 10.45 40.20 A 10.45

=
20  + 12   j 20 
32   j 20 
37.74   32.01
R4 0 + Z
= 1.07 A 42.46
I4 =
CHAPTER 16
=
201
13.
14.
R3 + R4 = 2.7 k + 4.3 k = 7 k
R = 3 k  7 k = 2.1 k
Z = 2.1 k  j10 
(40 k  0)(20 mA 0)
40 k  + 2.1 k   j10 
= 19 mA +0.014 as expected since R1  Z
(CDR)
I (of 10  cap.) =
(CDR)
I4 =
(3 k  0)(19 mA 0.014) 57 mA 0.014
=
3k + 7 k
10
= 5.7 mA 0.014
P = I2R = (5.7 mA)2 4.3 k = 139.71 mW
Z = X C2   90  R1 0 = 2  90  1  0
2    90
2    90

1  j2
2.236   63.435
= 0.894  26.565
Z = X L2 90 + Z = +j8  + 0.894  26.565
=
= +j8  + (0.8   j4 )
= 0.8  + j4 = 4.079  78.69
IXL =
2
X C1   90I
X C1   90  Z

(2   90)(0.5 A 0) 1 A   90

 j 2   (0.8   j 4 )
0.8  j 2
1 A   90
= 0.464 A  158.99
2.154 68.199
X C2   90 I X C
(2    90)(0.464 A   158.99) 0.928 A   248.99
2
=
=
I1 =
X C2   90 + R1
 j2  + 1 
2.236   63.435
=
= 0.42 A 174.45
202
CHAPTER 16
Chapter 17
1.

2.
Z = −j5  + 2  0  5  90 = −j5  + 1.72  + j0.69  = 4.64  −68.24
E
60 V 30
I=
= 12.93 A 98.24
=
Z 4.64    68.24
3.
Z = 10  0  6  90 = 5.15  59.04
E = IZ = (2 A 120)(5.15  59.04)
= 10.30 V 179.04
4.
a.
I=
b.
V = (hI)(R) = (50 I)(50 kΩ) = 2.5  106 I
Z = 50 k 0
5.
V
16 V
= 4  103 V
R
4  103
Z = 4 k 0
=
Clockwise mesh currents:
E  I1Z1  I1Z2 + I2Z2 = 0
I2Z2 + I1Z2  I2Z3  E2 = 0
────────────────────
[Z1 + Z2]I1  Z2I2 = E1
Z2I1 + [Z2 + Z3]I2 = E2
──────────────────
Z 2
E1
 E2
 Z 2 + Z3 
I R1 = I1 =
 Z1 + Z 2  Z 2
Z 2
6.
Z1 = R1 0 = 4  0
Z2 = XL 90 = 6  90
Z3 = XC 90 = 8  90
E1 = 10 V 0, E2 = 40 V 60
=
 Z 2 + Z3  E1  Z 2 E2 = 5.15 A 24.5
Z1Z 2 + Z1Z3 + Z 2 Z 3
 Z 2 + Z3 
By interchanging the right two branches, the general configuration of Problem 5 will result and
I50Ω = I1 =
 Z 2  Z3  E1  Z 2E2
Z1Z 2 + Z1Z3 + Z 2 Z 3
= 0.44 A 143.48
CHAPTER 17
Z1 = R1 = 50  0
Z2 = XC 90 = 60  90
Z3 = XL 90 = 20  90
E1 = 5 V 30, E2 = 20 V 0
203
7.
a.
Z1 = 12  + j12  = 16.971  45
Z2 = 3  0
Z3 = j1 
E1 = 20 V 50
E2 = 60 V 70
E3 = 40 V 0
I1[Z1 + Z2]  Z2I2 = E1  E2
I2[Z2 + Z3]  Z2I1 = E2  E3
───────────────────
(Z1 + Z2)I1  Z2I2 = E1  E2
Z2I1 + (Z2 + Z3)I2 = E2  E3
─────────────────────
Using determinants:
(E  E2 )(Z 2 + Z 3 ) + Z 2 (E 2  E3 )
= 2.55 A 132.72
I R1 = I1 = 1
Z1Z 2 + Z1Z3 + Z 2 Z 3
8.
Clockwise mesh currents:
E1  I1Z1  I1Z2 + I2Z2 = 0
I2Z2 + I1Z2  I2Z3  I2Z4 + I3Z4 = 0
I3Z4 + I2Z4  I3Z5  E2 = 0
─────────────────────────
 Z2 I 2
[Z1 + Z 2 ]I1
Z2
I1 + [Z 2 + Z 3 + Z 4] I 2
Z1 = 4  + j3 , Z2 = j1 
Z3 = +j6 , Z4 = j2 
Z5 = 8 
E1 = 60 V 0, E2 = 120 V 120
+0
= E1
 Z4 I3 = 0
 Z 4 I 2 + [ Z 4 + Z 5]I 3 =  E 2
0
───────────────────────────────────
I R1 = I 3 =
 Z 2Z 4  E1 +  Z 22   Z1 + Z 2  Z 2 + Z3 + Z 4 E2
 Z1 + Z 2  Z 2 + Z3 + Z 4  Z 4 + Z5   Z1 + Z 2 Z 24   Z 4 + Z5 Z 22
= 13.07 A   33.71°
204
CHAPTER 17
9.
Z1 = 15  0, Z2 = 15  0
Z3 = j10  = 10  90
Z4 = 3  + j4  = 5  53.13
E1 = 220 V 0
E2 = 100 V 90
I1(Z1 + Z3)  I2Z3  I3Z1 = E1
I2(Z2 + Z3)  I1Z3  I3Z2 = E2
I3(Z1 + Z2 + Z4)  I1Z1  I2Z2 = 0
───────────────────────
 I3Z1
= E1
I1(Z1 + Z3)  I2Z3
+ I2(Z2 + Z3)  I3Z2
= E2
I1Z3
 I2Z2
+ I3(Z1 + Z2 + Z4) = 0
I1Z1
───────────────────────────────────
Applying determinants:
I3 =

( Z1 + Z 3)( Z 2 )E 2  Z1Z 3E 2 + E1  Z 2 Z 3 + Z1( Z 2 + Z 3) 

( Z1 + Z 3) ( Z 2 + Z 3)( Z1 + Z 2 + Z 4 )  Z 22 + Z 3  Z 3( Z1 + Z 2 + Z 4 )  Z1Z 2   Z1   Z 2 Z 3  Z1( Z 2 + Z 3) 
= 48.33 A 77.57

or I3 = E1 E 2 if one carefully examines the network!
Z4
10.
Z1 = 5  0, Z2 = 5  90
Z3 = 4  0, Z4 = 6  90
Z5 = 4  0, Z6 = 6  + j8 
E1 = 20 V 0, E2 = 40 V 60
I1(Z1 + Z2 + Z4)  I2Z2  I3Z4 = E1
I2(Z2 + Z3 + Z5)  I1Z2  I3Z5 = E2
I3(Z4 + Z5 + Z6)  I1Z4  I2Z5 = 0
─────────────────────────
 Z2I2
 Z4I3 = E1
(Z1 + Z2 + Z4) I1
 Z5I3 = E2
Z2I1 + (Z2 + Z3 + Z5)I2
 Z5I2 + (Z4 + Z5 + Z6)I3 = 0
Z4I1
─────────────────────────────────────────
Using Z = Z1 + Z2 + Z4, Z = Z2 + Z3 + Z5, Z = Z4 + Z5 + Z6 and determinants:
E1 (Z  Z   Z 52 )  E 2 (Z 2 Z  + Z 4 Z5 )
I R1 = I1 =
Z  (Z  Z   Z52 )  Z 2 (Z 2 Z  + Z 4 Z5 )  Z 4 (Z 2 Z5 + Z 4 Z  )
= 3.04 A 169.12°
CHAPTER 17
205
11.
Z1 = 10  + j20 
Z3 = 80  0
Z5 = 15  90
Z7 = 5  0
E1 = 25 V 0
Z2 = j20 
Z4 = 6  0
Z6 = 10  0
Z8 = 5 Ω  j20 
E2 = 75 V 20
I1(Z4 + Z6 + Z7)  I2Z4  I4Z6 = E1
I2(Z1 + Z2 + Z4)  I1Z4  I3Z2 = 0
I3(Z2 + Z3 + Z5)  I2Z2  I4Z5 = E2
I4(Z5 + Z6 + Z8)  I1Z6  I3Z5 = 0
─────────────────────────
 Z4 I2
+0
 Z6I4 = E1
(Z4 + Z6 + Z7) I1
 Z2I3
+0 =0
Z4I1 + (Z1 + Z2 + Z4)I2
 Z5I4 = E2
0
 Z2 I2 + (Z2 + Z3 + Z5)I3
+0
 Z5I3 + (Z5 + Z6 + Z7)I4 = 0
Z6I1
────────────────────────────────────────────────────
Applying determinants:
I R1 = I 80 = 0.68 A 162.9
12.
Z1 = 5 k 0
Z2 = 10 k 0
Z3 = 1 k + j4 k = 4.123 k 75.96
I1(Z1 + Z2)  Z2I2 = 28 V
I2(Z2 + Z3)  Z2I1 = 0
──────────────────
(Z1 + Z2)I1  Z2I2 = 28 V
Z2I1 + (Z2 + Z3)I2 = 0
───────────────────
Z 2 28 V
= 3.17  103 V 137.29
IL = I2 =
Z1Z 2 + Z1Z 3 + Z 2 Z3
206
CHAPTER 17
Source Conversion:
E = (I θ)(Rp 0)
= (50 I)(40 k 0)
= 2  106 I 0
Z1 = Rs = Rp = 40 k 0
Z2 = j0.2 k
Z3 = 8 k 0
Z4 = 4 k 90
13.
I1(Z1 + Z2 + Z3)  Z3I2 = E
I2(Z3 + Z4)  Z3I1 = 0
────────────────────
(Z1 + Z2 + Z3)I1  Z3I2 = E
Z3I1 + (Z3 + Z4)I2 = 0
────────────────────
Z3 E
= 42.91 I 149.31
IL = I2 =
(Z1 + Z 2 + Z3 )(Z3 + Z 4 )  Z 23
14.
6Vx  I1 1 k  10 V 0 = 0
10 V0  I2 4 k  I2 2 k = 0
───────────────────────
Vx = I2 2 k
I1 1 k + I2 12 k = 10 V 0
I2 6 k = 10 V 0
──────────────────────
10 V 0
I2 = I 2k =
= 1.67 mA 0 = I2kΩ
6 k
I1 1 k + (1.667 mA 0)(12 k) = 10 V 0
I1 1 k + 20 V 0 = 10 V 0
I1 1 k = 10 V 0
10 V 0
= 10 mA 0
I1 = I1k =
1k
CHAPTER 17
207
E1 = 5 V 0
E2 = 20 V 0
Z1 = 2.2 k 0
Z2 = 5 k 90
Z3 = 10 k 0
I = 4 mA 0
15.
E1  I1Z1  Z2(I1  I2) = 0
Z2(I2  I1) + E2  I3Z3 = 0
───────────────────
I3  I2 = I
Substituting, we obtain:
I1(Z1 + Z2)  I2Z2 = E1
I1Z2  I2(Z2 + Z3) = IZ3  E2
────────────────────
Determinants:
I1 = 1.39 mA 126.48, I2 = 1.341 mA 10.56, I3 = 2.693 mA 174.8
I10kΩ = I3 = 2.69 mA 174.8
Z1 = 1 k 0
Z2 = 4 k + j6 k
E = 10 V 0
16.
Z1(I2  I1) + E  I3Z3 = 0
I1 = 6 mA 0, 0.1 Vs = I3  I2, Vs = (I1  I2)Z1
─────────────────────────────────
Substituting:
(1 k)I2 + (4 k + j6 k)I3 = 16 V 0
I3 = 0.6 V 0
(99 )I2 +
────────────────────────────
Determinants:
I3 = I6 kΩ = 1.38 mA 56.31
208
CHAPTER 17
17.
Z1 = 4 Ω 0
Z2 = 5 Ω 90
Z3 = 2 Ω 90
I1 = 3 A 0
I2 = 5 A 30
I1 = I3 + I4
1
1 
1 
V1 V1  V 2
+
 V 1  +   V 2   = I1
Z1
Z2
 Z1 Z 2 
 Z2 
or V1[Y1 + Y2]  V2[Y2] = I1
I1 =
I4 = I5 + I2
1
1 
1
V1  V 2 V 2
=
+ I 2  V 2  +   V1   =  I 2
Z2
Z3
 Z2 
 Z 2 Z3 
or V2[Y2 + Y3]  V1[Y2] = I2
 Y2V2 = I1
[Y1 + Y2]V1
Y2V1 + [Y2 + Y3]V2 = I2
──────────────────────
[Y2 + Y3 ] I1  Y2 I 2
V1 =
= 14.68 V 68.89
Y1 Y2 + Y1 Y3 + Y2 Y3
[Y1 + Y2 ] I 2 + Y2 I1
= 12.97 V 155.88
V2 =
Y1 Y2 + Y1 Y3 + Y2 Y3
18.
Z1 = 3  + j4  = 5 53.13
Z2 = 2  0
Z3 = 6  0  8  90
= 4.8  36.87
I1 = 0.6 A 20
I2 = 4 A 80
0 = I1 + I3 + I4 + I2

0 = I1 + V1 + V1 V 2 + I 2
Z1
Z2
1
1 
1 
V1  +   V 2   =  I1  I 2
 Z1 Z 2 
 Z2 
or
V1[Y1 + Y2]  V2[Y2] = I1  I2
──────────────────────
I2 + I4 = I5
V  V2 V2
=
I2 + 1
Z2
Z3
CHAPTER 17
209
 1
 1 
1 
V2 
+
  V1   = + I 2
 Z2 
 Z 2 Z3 
or
V2[Y2 + Y3]  V1[Y2] = I2
──────────────────
and
[Y1 + Y2]V1  Y2V2 = I1  I2
Y2V1 + [Y2 + Y3]V2 = I2
Applying determinants:
[ Y 2 + Y 3][I1 + I 2] + Y 2I 2
= 5.12 V 79.36
Y1Y 2 + Y1Y 3 + Y 2 Y 3
Y1I 2  I1Y 2
V2 =
= 2.71 V 39.96
Y1Y 2 + Y1Y 3 + Y 2 Y 3
V1 =
19.
Z1 = 5  0
Z2 = 6  90
Z3 = 4  90
Z4 = 2  0
E = 30 V 50
I = 0.04 A 90
I1 = I2 + I3
1
1
1 V
E1  V1 = V1 + (V1  V 2) 
+   2 = E1
V1  +
Z1
Z2
Z3
Z1
 Z1 Z 2 Z3  Z 3
or V1[Y1 + Y2 + Y3]  Y3V2 = E1Y1
I3 + I = I4
V1  V 2
+I=
Z3
1
1 
V2
 V 2  +   V1 = I
Z4
 Z3 Z4  Z3
or V2[Y3 + Y4]  V1Y3 = I
resulting in
V1[Y1 + Y2 + Y3]  V2Y3 = E1Y1
V1[Y3] + V2[Y3 + Y4] = +I
──────────────────────
Using determinants:
V1 = 19.86 V 43.8 and V2 = 8.94 V 106.9
210
CHAPTER 17
20.
Z1 = 10  0
Z2 = 10  0
Z3 = 4  90
Z4 = 2  0
Z5 = 8  90
E = 50 V 120
I = 0.8 A 70
I1 = I2 + I5
1
1
E  V1
(V1  V 2)
1
1
1
1 E
V
V  V2
= 1 +
+ 1
 V1  +
+
+   V2 
+
=
Z1
Z2
Z5
Z3
Z 5  Z1
 Z1 Z 2 Z 3 Z 5 
 Z3
or V1[Y1 + Y2 + Y3 + Y5]  V2[Y3 + Y5] = E1Y1
I3 + I5 = I4 + I
1
1
1
1
1
V1  V 2 V1  V 2 V 2
+
=
+ I  V2  +
+   V1 
+
 = I
Z3
Z5
Z4
 Z3 Z 4 Z5 
 Z3 Z5 
or V2[Y3 + Y4 + Y5]  V1[Y3 + Y5] = I
resulting in
V1[Y1 + Y2 + Y3 + Y5]  V2[Y3 + Y5] = E1Y1
V1[Y3 + Y5] + V2[Y3 + Y4 + Y5] = I
───────────────────────────────
Applying determinants:
V1 = 19.78 V 132.48 and V2 = 13.37 V 98.78
21.
Z1 = 15  0
Z2 = 10  90
Z3 = 15  0
Z4 = 3  + j4 
1
1
1 1
1
+ 
V2  +
V1 
V3 = 0
Z3
 Z1 Z 2 Z 3  Z1
 1
1
1 
1
1

+
+
 220 V 0 
100 V 90 = 0
V2 

15 
15   j10  15   15 
3
3
V 2 133.34  10  j100  10  14.67  j 6.67
V2 
16.05 V 24.55
= 96.30 V 12.32
166.67  103 36.37
V1 = E1 = 220 V 0, V3 = E2 = 100 V 90
CHAPTER 17
211
22.
E1 = 25 V 0
E2 = 75 V 20
Z1 = 10  + j20 
Z2 = 6  0
Z3 = 5  0
Z4 = 20  90
Z5 = 10  0
Z6 = 80  0
Z7 = 15  90
Z8 = 5   j20 Ω



V1: V1 V 2 + V1 V 4 + V1 E1 = 0
Z1
Z2
Z3


 
V2: V 2 V1 + V 2 V 4 + V 2 E 2 V 3 = 0
Z1
Z4
Z6
+


V3: V 3 E 2 V 2 + V 3 V 4 + V 3 = 0
Z6
Z7
Z8



V4: V 4 V1 + V 4 V 2 + V 4 V 3 + V 4 = 0
Z2
Z4
Z7
Z5
───────────────────────
Rearranging:
 1
1
1 
V
V
E
V1  +
+   2 4 = 1
Z1 Z 2
Z3
 Z1 Z 2 Z 3 
 1
1
1  V1 V 4 V 3


= E2
V2  +
+
 
Z6
Z1 Z 4 Z 6
 Z1 Z 4 Z 6 
 1
1
1 
V
V
E
V3 
+
+   2 4 =  2
Z6 Z 7
Z6
 Z 6 Z 7 Z8 
 1
1
1
1 
V V
V
V4 
+
+
+   1 2 3 =0
 Z 2 Z 4 Z7 Z5  Z 2 Z 4 Z 7
Setting up and then using determinants:
V1 = 14.62 V 5.86, V2 = 35.03 V 37.69
V3 = 32.4 V 73.34, V4 = 5.67 V 23.53
212
CHAPTER 17
23.
1
4 Ω 0°
= 0.25 S 0
1
Y2 =
1 Ω 90
= 1 S 90
1
Y3 =
5  0
= 0.2 S 0
1
Y4 =
4    90
= 0.25 S 90
1
Y5 =
8 Ω 90°
= 0.125 S 90
I1 = 2 A 30
I2 = 3 A 150
Y1 =
V1[Y1 + Y2]  Y2V2 = I1
V2[Y2 + Y3 + Y4]  Y2V1  Y4V3 = I2
V3[Y4 + Y5]  Y4V2 = I2
───────────────────────────
 Y2 V2
+ 0 = I1
[Y1 + Y2]V1
 Y4 V3 = I2
Y2V1 + [Y2 + Y3 + Y4]V2
0
 Y4 V2 + [Y4 + Y5]V3 = I2
────────────────────────────────────
2
I1  (Y 2 + Y 3 + Y 4)(Y 4 + Y 5)  Y 4   I 2[Y 2 Y5]
V1 =
 Y1 + Y 2 (Y 2 + Y3 + Y 4)(Y 4 + Y 5)  Y 24   Y 22(Y 4 + Y5) = Y 
= 5.74 V 122.76
( + )
( + )
V2 = I1Y 2 Y 4 Y 5 I 2 Y 5 Y1 Y 2 = 4.04 V 145.03
Y
2
I 2  (Y1 + Y 2)(Y 3 + Y 4)  Y 2   Y 2 Y 4I1
V3 = 
= 25.94 V 78.07
Y
CHAPTER 17
213
24.
1
4 Ω 0°
= 0.25 S 0
1
Y2 =
6  0
= 0.167 S 0
1
Y3 =
8  0
= 0.125 S 0
1
Y4 =
2    90
V1[Y1 + Y2 + Y3]  Y2V2  Y3V3 = I1
= 0.5 S 90
V2[Y2 + Y4 + Y5]  Y2V1  Y4V3 = 0
1
Y5 =
V3[Y3 + Y4 + Y6]  Y3V1  Y4V2 = I2
5  90
───────────────────────────
= 0.2 S 90
 Y2V2
 Y3V3 = I1
[Y1 + Y2 + Y3]V1
1
 Y4V3 = 0
Y2V1 + [Y2 + Y4 + Y5]V2
Y6 =
4  90
 Y4V2 + [Y3 + Y4 + Y6]V3 = I2
Y3V1
= 0.25 S 90
──────────────────────────────────────────
I1 = 4 A 0
I2 = 6 A 90
I  ( Y + Y + Y )( Y + Y + Y )  Y   I  Y Y + Y ( Y + Y + Y ) 
V1 =
Y = ( Y  Y + Y )  ( Y + Y + Y )( Y  Y + Y )  Y   Y  Y ( Y + Y + Y )+ Y Y   Y  Y Y + Y ( Y + Y + Y ) 
Y1 =
2
1
2
4
5
3
4
6
4
2
2
4
3
4
6
3
4
5
2

1
2
3
2
4
5
3
4
6
4
2
2
3
3
4
3
2
4
3
2
4
5
= 15.13 V 1.29
I1 (Y 2)(Y 3 + Y 4 + Y 6)+ Y 3Y 4  +I 2  Y 4(Y1 + Y 2 + Y 3)  Y 2 Y 3
V2 =
= 17.24 V 3.73
Y
2
I1 (Y3)(Y 2 + Y 4 + Y 5) + Y 2 Y 4  + I 2  Y 2 - (Y1 + Y 2 + Y 3)(Y 2 + Y 4 + Y5) 
V3 =
Y
= 10.59 V 0.11
25.
Left node:
V1
 Ii   Io
4Ix = Ix + 5 mA 0 +
V1  V 2
2 k
Right node: V2
 Ii   I o
8 mA 0 =
V2
V  V1
+ 2
+ 4I x
1k 
2 k
V1
4 k    90
Rearrange, reduce and 2 equations with 2 unknowns result:
V1[1.803 123.69] + V2 = 10
V1[2.236 116.57] + 3 V2 = 16
──────────────────────
Determinants:
V1 = 4.37 V 128.66
V2 = V1k = 2.25 V 17.63
Insert Ix =
214
CHAPTER 17
Z1 = 1 kΩ 0
Z2 = 2 kΩ 90
Z3 = 3 kΩ 90
I1 = 12 mA 0
I2 = 4 mA 0
E = 10 V 0
26.
 Ii   Io
V1 V 2
+
+ I2
Z1 Z 3
and V1 + V 2 = I1  I2
Z1 Z 3
with V2  V1 = E
0 = I1 +
Substituting and rearranging:
1
1
E
V1  +  = I1  I2 
Z3
 Z1 Z 3 
and solving for V1:
V1 = 15.4 V  178.2
with V2 = VC = 5.41 V  174.87
27.
Left node: V1
 Ii   Io
2 mA 0 = 12 mA 0 +
and 1.5 V1  V2 = 10
Right node: V2
 Ii   I o
V1
V  V2
+ 1
2 k
1k
V 2  V1 V 2  6 V x

1k
3.3 k 
and 2.7 V1  3.7 V2 = 6.6
0 = 2 mA 0 +
Using determinants:
CHAPTER 17
V1 = V2k = 10.67 V 0 = 10.67 V 180
V2 = 6 V 0 = 6 V 180
215
Z1 = 2 k 0
Z2 = 1 k0
Z3 = 1 k 0
I = 5 mA 0
28.
V1
V
+ 3I1 + 2
Z1
Z3
with I1 = V1 V 2
Z2
and V2  V1 = 2Vx = 2V1 or V2 = 3V1
V1: I =
Substituting will result in:
or
and
with
29.
1
1
3
3
V1  +  + 3 V1    = I
 Z1 Z 2 
 Z3 Z 2 
1
6
3
+  =I
V1  
 Z1 Z 2 Z 3 
V1 = Vx = 2 V 0
V2 = 6 V 0
Ei 
= 1  103 Ei
R1 0
1
Y1 =
= 0.02 mS 0
50 k
1
= 1 mS 0
Y2 =
1k
Y3 = 0.02 mS 0
I2 = (V1  V2)Y2
I1 =
V1(Y1 + Y2)  Y2V2 = 50I1
V2(Y2 + Y3)  Y2V1 = 50I2 = 50(V1  V2)Y2 = 50Y2V1  50Y2V2
────────────────────────────────────────────
(Y1 + Y2)V1  Y2V2 = 50I1
51Y2V1 + (51Y2 + Y3)V2 = 0
───────────────────────
(50)(51)Y2 I1
= 2451.92 Ei
VL = V2 =
(Y1 + Y2 )(51Y2 + Y3 )  51Y22
216
CHAPTER 17
30.
a.
yes
b.
Z1 = 5 k 0, Z2 = 8 k 0
Z3 = 2.5 k 90, Z4 = 4 k 90
Z5 = 5 k 90, Z6 = 1 k 0
Z1 = Z 2
Z3
Z4
3
8  103 0
5  10 0
=
2.5  103 90
4  103 90
2 90 = 2 90 (balanced)
I1[Z1 + Z3 + Z6]  Z1I2  Z3I3 = E
I2[Z1 + Z2 + Z5]  Z1I1  Z5I3 = 0
I3[Z3 + Z4 + Z5]  Z3I1  Z5I2 = 0
───────────────────────
[Z1 + Z3 + Z6]I1
 Z1I2
 Z3I3 = E
 Z5I3 = 0
Z1I1 + [Z1 + Z2 + Z5]I2
 Z5I2 + [Z3 + Z4 + Z5]I3 = 0
Z3I1
───────────────────────────────────────
I2 =
I3 =
E Z1( Z 3 + Z 4 + Z 5 ) + Z 3 Z 5 
Z  = ( Z1 + Z 3 + Z 6 )[( Z1 + Z 2 + Z 5 )( Z 3 + Z 4 + Z 5 )  Z 5 ]  Z 1[ Z 1( Z 3 + Z 4 + Z 5 )  Z 3 Z 5 ]  Z 3[ Z 1 Z 5 + Z 3 ( Z1 + Z 2 + Z 5 )]
2
E Z1Z5 + Z 3 (Z1 + Z 2 + Z5 ) 
I Z 5 = I2  I3 =
CHAPTER 17
Z
E Z1Z 4  Z 3 Z 2 
Z
=
E  20  106 90  20  106 90 
Z
=0A
217
c.
V1[Y1 + Y2 + Y6]  Y1V2  Y2V3 = I
V2[Y1 + Y3 + Y5]  Y1V1  Y5V3 = 0
V3[Y2 + Y4 + Y5]  Y2V1  Y5V2 = 0
─────────────────────────
 Y1V2
 Y2V3 = I
[Y1 + Y2 + Y6]V1
 Y5V3 = 0
Y1V1 + [Y1 + Y3 + Y5]V2
 Y5V2 + [Y2 + Y4 + Y5]V3 = 0
Y2V1
─────────────────────────────────────────
V2 =
V3 =
I Y1( Y 2 + Y 4 + Y 5 ) + Y 2 Y 5 
10 V 0
I = Es =
1 k  0
Rs
= 10 mA 0
1
Y1 =
5 k   0
= 0.2 mS 0
1
Y2 =
8 k  0
= 0.125 mS 0
1
Y3 =
2.5 k  90
= 0.4 mS 90
1
Y4 =
4 k  90
= 0.25 mS 90
1
Y5 =
5 k    90
= 0.2 mS 90
1
Y6 =
1 k  0
V2 = 1 mS 0
Y  = ( Y1 + Y 2  Y 6 )[( Y 1 + Y 3 + Y 5)( Y 2 + Y 4 + Y 5)  Y 5 ]  Y1[ Y1( Y 2 + Y 4 + Y 5 )  Y 2 Y 5]  Y 2[ Y 1Y 5 + Y 2( Y1+ Y 3 + Y 5 )]
2
I Y1Y5 + Y2 (Y1 +Y3 +Y5 )
Y
V Z 5 = V2  V3 =
I Y1Y4  Y4 Y3 
=
I  0.05  103   90  0.05  103   90 
Y
Y
=0V
31.
a.
Z1 = Z 2
Z3
Z4
3
4  10 0
4  103 0
?
3
4  10 90
4  103   90
1 90  1 90 (not balanced)
b.
The solution to 26(b) resulted in
E(Z1Z 5  Z 3 (Z1  Z 2  Z5 )
I3 = I X C =
Z
where
and
and
218
ZΔ = (Z1 + Z3 + Z6)[(Z1 + Z2 + Z5)(Z3 + Z4 + Z5)  Z 52 ]
 Z1[Z1(Z3 + Z4 + Z5)  Z3Z5]  Z3[Z1Z5 + Z3(Z1 + Z2 + Z5)]
Z1 = 5 k 0, Z2 = 8 k 0, Z3 = 2.5 k 90
Z4 = 4 k 90, Z5 = 5 k 90, Z6 = 1 k 0
I X C = 1.76 mA 71.54
CHAPTER 17
c.
The solution to 26(c) resulted in
I Y1Y5 + Y2 (Y1 +Y3 +Y5 )
V3 = VX C =
Y
YΔ = (Y1 + Y2 + Y6)[(Y1 + Y3 + Y5)(Y2 + Y4 + Y5)  Y52 ]
 Y1 [Y1(Y2 + Y4 + Y5) + Y2Y5]
 Y2[Y1Y5 + Y2(Y1 + Y3 + Y5)]
Y1 = 0.2 mS 0, Y2 = 0.125 mS 0, Y3 = 0.4 mS 90
Y4 = 0.25 mS 90, Y5 = 0.2 mS 90
where
with
Y6 = 1 mS 0, I = 10 mA 0
V3 = VX C = 7.03 V 18.46
Source conversion:
and
32.
Z1Z4 = Z3Z2


(R1  jXC) Rx + jX Lx = R3R2

XC =

1
1
= 1 k
=
3
C (10 rad/s)(1  F)
(1 k  j1 k) Rx + jX Lx = (0.1 k)(0.1 k) = 10 k
and Rx + jX Lx =
 Rx = 5 , Lx =
33.
10  103 
10  103
= 5 Ω + j5 Ω
=
1  103  j1  103 1.414  103   45
X Lx

=
5
= 5 mH
10 rad/s
3
1
1
1
=
= k
C (1000 rad/s)(3  F) 3
1
Z1 = R1  X C1 90 = (2 k 0)   k 90 = 328.8  80.54
3
Z2 = R2 0 = 0.5 k 0, Z3 = R3 0 = 4 k 0
Z4 = Rx + j X Lx = 1 k + j6 k
X C1 =
Z1 = Z 2
Z3
Z4
328.8    80.54
0.5 k 0
?
4 k 0
6.083  80.54
82.2  10380.54  82.2  103 80.54 (balanced)
34.
Apply Eq. 17.6.
CHAPTER 17
219
35.
For balance:
R1(Rx + j X Lx ) = R2(R3 + j X L3 )
R1Rx + jR1 X Lx = R2R3 + jR2 X L3
R2 R3
R1
R1 X Lx = R2 X L3 and R1ωLx = R2ωL3
 R1Rx = R2R3 and Rx =
so that Lx =
36.
R2 L3
R1
Z1 = 8  90 = j8 
Z2 = 4  90 = +j4 
Z3 = 8  90 = +j8 
Z4 = 6  90 = j6 
Z5 = 5  0
a.
Z1Z 2
= 5  38.66
+
Z1 Z 2 + Z5
Z1Z 5
= 6.25  51.34
Z7 =
Z1 + Z 2 + Z5
Z 2 Z5
Z8 =
= 3.125  128.66
Z1 + Z 2 + Z5
Z = Z7 + Z3 = 3.9  + j3.12  = 4.99  38.66
Z = Z8 + Z4 = 1.95   j3.56  = 4.06  118.71
Z  Z = 10.13  67.33= 3.90   j9.35 
ZT = Z6 + Z  Z = 7.80   j6.23  = 9.98  38.61
E
120 V 0
=
= 12.02 A 38.61
I=
ZT 9.98    38.61
Z6 =
37.
220
ZY = Z  =
3
12   j 9 
= 4   j3 
3
CHAPTER 17
ZT = 2  + 4  + j3  + [4   j3  + j3 ]  [4   j3  + j3 ]
= 6   j3  + 2 
= 8   j3  = 8.544  20.56
E
60 V 0
I=
= 7.02 A 20.56
=
8.544    20.56
ZT
ZΔ = 3ZY = 3(3  90) = 9  90
Z = 9  90  (12   j16 )
= 9  90  20  53.13
= 12.96  67.13
38.
2Z 2
2
2
= Z = [12.96  67.13] = 8.64  67.13
3
Z + 2Z 3
E
100 V 0
I=
= 11.57 A 67.13
=
ZT 8.64  67.13
ZT = Z  2Z =
39.
ZΔ = 3ZY = 3(5 ) = 15 
Z1 = 15  0  5  90
= 4.74  71.57
Z2 = 15  0  6  90
= 5.57  68.2 = 2.07  + j5.17 
Z3 = Z1 = 4.74  71.57
= 1.5   j4.5 
ZT = Z1 (Z2 + Z3) = (4.74  71.57)  (2.07  + j5.17  + 1.5   j4.5 )
= (4.74  7.57)  (3.63  10.63)
= 2.71  23.87
E
100 V 0
=
= 36.9 A 23.87
I=
ZT 2.71    23.87
CHAPTER 17
221
Chapter 18
1.
Z1 = 3  0, Z2 = 8  90, Z3 = 6  90
Z2  Z3 = 8  90  6  90 = 24  90
30 V 30
E1
=
= 1.24 A 112.875
Z1 + Z 2  Z 3 3   j 24 
Z3 I
(6    90)(1.24 A 112.875)
= 3.72 A 67.125
I =
=
2  90
Z 2 + Z3
I=
Z1 Z2 = 3  0  8  90 = 2.809  20.556
60 V 10
E2
=
I=
Z3 + Z1  Z 2  j 6  + 2.630  + j 0.986 
= 10.597 A 72.322
I =
Z1 I
(3  0)(10.597 A 72.322)
= 3.721 A 2.878
=
Z1 + Z 2
3  + j8 
I L1 = I + I = 3.72 A 67.125 + 3.721 A 2.878
= 1.446 A  j3.427 A + 3.716 A + j0.187 A
= 5.162 A  j3.24 A
= 6.09 A 32.12
2.
222
Z1 = 8  90, Z2 = 5 Ω 90
I = 0.3 A 60, E = 10 V 0
Z2 I
(8  90)(0.3 A 60) 2.4 A 150
I =
=

Z1 + Z 2
+ j8   j 5 
3 90
= 0.8A 60
E
10 V 0
10 A 0
I =

=
Z1 + Z 2 + j8   j 5 
3 90
= 3.33 A 90
IC = I  I
= 0.8 A 60  3.33 A 90
= (0.4 A + j0.69 A) + j3.33 A
= 0.4 A + j4.02 A
= 4.04 A 84.32
CHAPTER 18
3.
E:
Z1 = 3  90, Z2 = 7  90
E = 10 V 90
Z3 = 6  90, Z4 = 4  0
Z = Z1 (Z3 + Z4)
= 3  90  (4   j6 )
= 3  90  7.21  56.31
= 4.33  70.56
ZE
V1 =
Z + Z 2
(4.33  70.56)(10 V 90)
=
(1.44  + j 4.08 )  j 7
43.3 V 160.56
=
= 13.28 V 224.31
3.26   63.75
13.28 V 224.31
I = V1 =
3  90
Z1
= 4.43 A 134.31
I:
CDR:
Z = Z3 + Z1 Z2
= j6  + 3  90  7  90
= j6  + 5.25  90
= j6  + j5.25 
= j0.75  = 0.75 90
Z4 I
(4  0)(0.6 A 120)
2.4 A 120
=
=
I3 =
4   j 0.75 
4.07   10.62
Z 4 + Z
= 0.59 A 130.62
Z2I3
(7    90)(0.59 A 130.62) 4.13 A 40.62
=
=
I =
 j7  + j3 
4   90
Z 2 + Z1
= 1.03 A 130.62
IL = I  I (direction of I)
= 4.43 A 134.31  1.03 A 130.62
= (3.09 A + j3.17 A)  (0.67 A + j0.78 A) = 2.42 A + j2.39 A
= 3.40 A 135.36
CHAPTER 18
223
4.
AC:
1
1
1


2 fC C (1000)(4.7  F)
= 212.77 
XL = 2fL = L = (1000)(47 mH)
= 47 
XC =
Z1 = 212.77  90, Z2 = 47   0, Z3 = 22  + j47  = 51.89  64.92
Z2 Z3 = 29.23  30.66
ZT = Z1 + Z2 Z3 = j212.77  + 25.14  + j14.91 
= 25.14   j197.86  = 199.45  82.76
Is =
E
20 V 60
= 0.1 A 142.76

ZT 199.45    82.76
Z3I S
(51.89  64.92)(0.1 A 142.76) 5.19 A 207.68


22   j 47   47 
83.49 34.26
Z3  Z 2
I = 62.16 mA  173.42
3
and i = 62.16  10 sin (1000t + 173.42)
I=
DC:
5V
5V

22   47  69 
= 72.46 mA
I=
i = 72.46 mA + 62.16  10
5.
sin (1000t + 173.42)
DC:
AC:
224
3
(6  0)(I )
6  + 3   j1 
(6  0)(4 A 0)
=
9   j1 
24 A 0
=
9.055  6.34
= 2.65 A 6.34
IC =
CHAPTER 18
VC = ICXC = (2.65 A 6.34)(1  90) = 2.65 V 83.66
= 12 V + 2.65 V 83.66
υC = 12 V + 3.75 sin(ωt  83.66)
6.
E = 20 V 0
Z1 = 10 k 0
Z2 = 5 k  j5 k
= 7.071 k 45
Z3 = 5 k 90
I = 5 mA 0
Z = Z1  Z2 = 10 k 0  7.071 k 45 = 4.472 k 26.57
(CDR)
ZI
(4.472 k   26.57)(5 mA 0) 22.36 mA   26.57
=
=
Z + Z 3
4 k  j 2 k + j5 k
5 36.87
= 4.472 mA 63.44
I =
Z = Z2  Z3
= 7.071 k 45  5 k 90
= 7.071 k 45
ZE
(7.071 k 45)(20 V 0) 141.42 V 45
=
=

Z + Z1
(5 k + j 5 k) + (10 k)
15.81 18.435
= 8.945 V 26.565
V  8.945 V 26.565
=
= 1.789 mA 63.435 = 0.8 mA  j1.6 mA
I =
5 k 90
Z3
I = I + I = (2 mA  j4 mA) + (0.8 mA  j1.6 mA) = 2.8 mA  j5.6 mA
= 6.26 mA 63.43
(VDR)
V =
7.
Z1 = 20 k 0
Z2 = 10 k 90
I = 2 mA 0
E = 10 V 0
I =
CHAPTER 18
Z1 (hI )
(20 k  0)(100)(2 mA 0)
=
= 0.179 A 26.57
Z1 + Z 2
20 k + j10 k
225
E
10 V 0
=
22.36 k  26.57
Z1 + Z 2
= 0.447 mA 26.57
IL = I  I (direction of I)
= 179 mA 26.57  0.447 mA 26.57
= 178.55 mA 26.57
I =
V:
8.
Z1 = 5 k 0, Z2 = 1 kΩ 90
Z3 = 4 k 0
V = 2 V 0, μ = 20
VL =
 Z 3(  V )
(4 k 0)(20)(2 V 0)
= 17.67 V 6.34
=
5 k  j1 k + 4 k
Z1 + Z 2 + Z 3
I:
CDR: I =
Z1I
Z1 + Z 2 + Z 3
(5 kΩ 0)(2 mA 0)
9.056 kΩ   6.34
= 1.104 mA 6.34
=
VL = IZ3 = (1.104 mA 6.34)(4 k 0) = 4.416 V 6.34
VL = VL + VL = 17.67 V 6.34  4.416 V 6.34 = 22.09 V 6.34
9.
Z1 = 20 k 0
Z2 = 5 k + j5 k
I =
Z1 (hI )
(20 k 0)(100)(1 mA 0)
=
= 78.45 mA 11.31
Z1 + Z 2
20 k + 5 k + j 5 k
V
(20)(10 V 0)
Z1 + Z 2
25.495 k 11.31
= 7.845 mA 11.31
I =
=
IL = I  I (direction of I)
= 78.45 mA 11.31  7.845 mA 11.31
= 70.61 mA 11.31
226
CHAPTER 18
Z1 = 2 k 0, Z2 = 2 k 0
VL = ILZ2
IL = hI + I = (h + 1)I
VL = (h + 1)IZ2
and by KVL: VL = IZ1 + E
V E
so that I = L
Z1
10.
V  E
VL = (h + 1)IZ2 = (h + 1)  L
 Z2
 Z1 
Subt. for Z1, Z2
VL = (h + 1)(VL  E)
VL(2 + h) = E(h + 1)
(h + 1)
51
VL =
E=
(20 V 53) = 19.62 V 53
(h + 2)
52
11.
I1:
I1 = 1 mA 0
Z1 = 2 k 0
Z2 = 5 k 0
KVL: V1  20 V  V = 0
I =
Z
V1
21 V
 I =
or V = 1 I
Z1
Z1
21
V1 = 21 V
V = I5Z2 = [I1  I]Z2
Z1
I = I1Z2  IZ2
21
Z

I 1 + Z 2  = I1Z 2
21


Z2
( 5 k  0)(1 mA 0)
and I =
[I1] =
= 0.981 mA 0
Z1
 2 k  0 
+ Z2

 + 5 k  0
21
21


CHAPTER 18
227
I2:
V1 = 20 V + V = 21 V
Z
V 21 V
 V = 1 I
I = 1 =
Z1
Z1
21
Z1
V
I5 =
=
I
Z 2 21 Z 2
I = I2  I5 = I2 
Z1
I
21 Z 2

Z 
I 1 + 1  = I2
 21 Z 2 
2 mA 0
I2
I =
=
= 1.963 mA 0
Z1
2 k
1+
1+
21(5 k)
21 Z 2
I = I + I = 0.981 mA 0 + 1.963 mA 0
= 2.94 mA 0
12.
E1:
10 V 0  I 10   I 2   4 Vx = 0
with Vx = I 10 
Solving for I:
10 V 0
= 192.31 mA 0
I=
52 
Vs = 10 V 0  I(10 ) = 10 V  (192.31 mA 0)(10  0) = 8.08 V 0
228
CHAPTER 18
I:
 Ii   Io
Vx 5 Vx
+
=
10  2 
5 A + 0.1 Vx + 2.5 Vx = 0
2.6 Vx = 5 A
5
V = 1.923 V
Vx = 
2.6
Vs = Vx = (1.923 V) = 1.923 V 0
Vs = Vs  Vs = 8.08 V 0 + 1.923 V 0 = 10 V 0
5 A 0 +
13.
ZTh:
Z1 = 3  0, Z2 = 4  90
E = 100 V 0
ZTh = Z1  Z2 = (3  0  4  90)
= 2.4  36.87 = 1.92  + j1.44 
ETh:
Z 2E
(4  90)(100 V 0)
=
Z 2 + Z1
5  53.13
= 80 V 36.87
ETh =
14.
ZTh:
ZTh = Z3 + Z1  Z2
= +j6 k + (2 k 0  3 k 90)
= +j6 k + 1.664 k 33.69
= +j6 k + 1.385 k j0.923 k
= 1.385 k + j5.077 k
= 5.26 k 74.74
ETh:
CHAPTER 18
ETh =
Z 2E
(3 k   90)(20 V 0)
=
Z 2 + Z1
2 k  j 3 k
=
60 V   90
= 16.64 V 33.69
3.606   56.31
229
15.
From #31. ZTh = Z1  Z2
ZTh = ZN = 21.31  32.2
ETh = IZ = IZTh
= (0.1 A 0)(21.31  32.12)
= 2.13 V 32.2
16.
From #31. ZTh = ZN = 6.81 Ω 54.23 = 3.98   j5.53 
Z1 = 2  0, Z3 = 8  90
Z2 = 4  90, Z4 = 10  0
E = 50 V 0
ETh = V2 + V4
Z2 E
V2 =
Z 2 + Z1  (Z3 + Z 4 )
( 4  90)(50 V 0)
=
+ j 4  + 2  0  (10   j8 )
= 47.248 V 24.7
V1 = E  V2 = 50 V 0  47.248 V 24.7 = 20.972 V 70.285
Z 4 V1
(10  0)(20.972 V   70.285)
=
= 16.377 V 31.625
V4 =
10   j8 
Z 4 + Z3
ETh = V2 + V4 = 47.248 V 24.7 + 16.377 V 31.625
= (42.925 V + j19.743 V) + (13.945 V  j8.587 V)
= 56.870 V + j11.156 V = 57.95 V 11.10
17.
ZTh:
Z1 = 10  0, Z2 = 8  90
Z3 = 8  90
ZTh = Z3 + Z1  Z2
= j8  + 10  0  8  90
= j8  + 6.247  51.34
= j8  + 3.902  + j4.878 
= 3.902   j3.122 
= 5.00  38.66
230
CHAPTER 18
ETh: Superposition:
(E1)
(8  90)(120 V 0)
10  + j8 
960 V 90
=
12.806 38.66
= 74.965 V 51.34
ETh =
(I)
ETh = VZ 2 + VZ3
= IZ3 + I(Z1  Z2)
= I(Z3 + Z1  Z2)
= (0.5 A 60)(j8  + 10  0  8  90)
= (0.5 A 60)(j8  + 3.902  + j4.878 )
= (0.5 A 60)(3.902 Ω  j3.122 Ω)
= (0.5 A 60)(4.997 Ω 38.663)
= 2.499 V 21.337
ETh = ETh + ETh
= 74.965 V 51.34 + 2.449 V 21.337
= (46.83 V + j58.538 V) + (2.328 V + j0.909 V)
= 49.158 V + j59.447 V = 77.14 V 50.41
18.
ZTh:
ZTh = Z = 10   j10  = 14.14  45
ETh:
CHAPTER 18
ETh = E  VZ
= 20 V 40  IZ
= 20 V 40  (0.6 A 90)(14.14  45)
= 20 V 40  8.484 V 45
= (15.321 V + j12.856 V)  (6 V + j6 V)
= 9.321 V + j6.856 V
= 11.57 V 36.34
231
19.
a.
AC:
ETh:
ETh =
1
  90  212.77    90
C
Z3 = 22  + L 90
= 22  + j47 
= 51.89   64
Z1 =
Z 3E
(51.89  64.92)(20 V 60)
= 6.21 V  207.36

Z3  Z1
22   j 47   j 212.77 
ZTh:
(212.77    90)(51.89  64.92)
 j 212.77   22   j 47 
= 66.04  57.36  = 35.62  + j55.61 
ZTh = Z1  Z2 =
DC:
ETh:
ETh = 5 V
RTh:
RTh = 22 
232
CHAPTER 18
b.
AC:
ETh
ZTh  R L
6.21 V  207.36
=
35.62   j 55.61   47 
6.21 V  207.36
=
82.62   j 55.61 
6.21 V 207.36
=
99.59  33.94
= 62.36 mA 173.42
I=
DC:
5V
5V

22   47  69 
= 72.46 mA
I=
i = 72.46 mA + 62.36  103 sin (1000t + 173.42)
matching the results of Problem 4.
20.
a.
ZTh:
ZTh = Z R1  Z R2 = 6  + 3  = 9 
DC:
ETh = 12 V
AC:
ETh = IZ R1 = (4 A  0)(6  0) = 24 V  0
ETh = 12 V + 24 V  0
(DC) (AC)
CHAPTER 18
233
b.
DC: VC = 12 V
ZC E
AC: VC =
ZC + Z RTh
(1    90)(24 V 0)
 j1  + 9 
24 V   90
=
9.055   6.34
VC = 2.65 V 83.66
=
υC = 12 V+ 2.65 V 83.66
= 12 V + 3.75 sin(ωt  83.66)
21.
a.
ZTh:
1 Z1 = 10 k 0
5 Z2 = 5 k  j5 k
= 7.071 k 45
ZTh = Z1  Z2 = (10 k 0)  (7.071 k 45) = 4.47 k 26.57
Source conversion:
E1 = (Iθ)(R10) = (5 mA 0)(10 k 0) = 50 V 0
ETh =
Z 2 (E + E1 )
Z 2 + Z1
(7.071 k    45)(20 V 0 + 50 V 0)
(5 k   j 5 k ) + (10 k )
(7.071 k    45)(70 V 0)
=
(15 k   j 5 k )
494.97 V   45
=
15.811   18.435
= 31.31 V 26.57
=
b.
I=
ETh
31.31 V   26.565
=
ZTh + Z L 4.472 k    26.565 + 5 k  90
31.31 V   26.565
31.31 V   26.565
=
4 k   j 2 k  + j5 k 
4 k  + j3 k 
31.31 V   26.565
=
= 6.26 mA 63.44
5 k  36.87
=
234
CHAPTER 18
22.
Z1 = 10 k  0
Z2 = 10 k  0
Z3 = 1 k  90
ZTh = Z3 + Z1  Z2 = 5 k  j1 k  5.1 k 11.31
ETh: (VDR)
23.
ETh =
Z 2 (20 V ) (10 k  0)(20 V )
= 10 V
=
Z 2 + Z1
10 k  + 10 k 
ZTh:
0 Z1 = 40 k 0
.2 Z 2 = 0.2 k 90
Z 3 = 5 k 0
ZTh = Z3  (Z1 + Z2) = 5 k 0  (40 k  j0.2 k) = 4.44 k 0.03
I =
Z1 (100 I )
Z1 + Z 2 + Z3
(40 k  0)(100 I )
45 k    0.255
= 88.89 I 0.255
=
ETh = IZ3 = (88.89 I 0.255)(5 k 0) = 444.45  103 I 0.26
24.
ZTh:
 ZTh = Z1 = 20 k 0
ETh:
ETh = (hI)(Z1)
= (100)(2 mA  0)(20 k  0)
= 4 kV 0
CHAPTER 18
235
E:
ETh = ETh + ETh
= 4 kV 0 + 10 V 0
= 3990 V 0
25.
ZTh:
Z1 = 5 k 0
Z2 = j1
ZTh = Z1 + Z2 = 5 k  j1 k
= 5.10 k 11.31
ETh:
ETh     V + VZ1 
=   V  IZ1
= (20)(2 V 0) − (2 mA 0)(5 k 0)
= −50 V 0
26.
ZTh:
Z1 = 20 k  0
Z2 = 5 k  0
ZTh = Z1 + Z2 = 25 k  0
ETh:
ETh = V  (hI)(Z1)
= (20)(10 V  0)  (100)(1 mA 0)(20 k  0)
= 1800 V 0
236
CHAPTER 18
27.
ETh: (Eoc)
hI = I
Z1 = 2 k 0
I = 0
and hI = 0
with Eoc = ETh = E = 20 V 53
Isc:
Isc = (h + 1)I
= (h + 1)(10 mA 53)
= 510 mA 53
ZTh =
28.
Eoc
20 V 53
= 39.22  0 (negative resistance)
=
I sc 510 mA 53
ETh:
Eoc = 21 V
Z1 = 5 k 0
V = I1Z1 = (1 mA 0)(5 k 0)
= 5 V 0
Eoc = ETh = 21(5 V 0)
= 105 V 0
V = I2Z1
= (2 mA 0)(5 k 0)
= 10 V 0
Eoc = ETh = V + 20 V = 21 V = 210 V 0
Isc:
Isc = I1
20 V = V  V = 0 V
and I = 0 A
 Isc = I2
Isc = Isc + Isc = 3 mA 0
Eoc = Eoc + Eoc = 315 V 0 = ETh
315 V 0
= 105 k 0
ZTh = Eoc =
3 mA 0
I sc
CHAPTER 18
237
29.
Eoc:
(ETh)
KVL: 6 Ix(2 k)  Ix(1 k) + 8 V 0  Ix(3.3 k) = 0
8 V 0
= 0.491 mA 0
Ix =
16.3 k 
Eoc = ETh = Ix(3.3 k) = 1.62 V 0
Isc:
8V
= 2.667 mA 0
3k
E
1.62 V 0
= 607.42  0
ZTh = oc =
2.667 mA 0
I sc
Isc =
30.
From Problem 13: ZN = ZTh = 1.92  + j1.44  = 2.4  36.87
I N:
Z1 = 3  0, Z2 = 4  90
E 100 V 0
Isc = IN =
=
Z1
3  0
= 33.33 A 0
238
CHAPTER 18
31.
Z1 = 20  + j20  = 28.284  45
Z2 = 68  0
ZN = Z1  Z2
= (28.284  45)  (68  0)
= 21.31  32.2
Isc = I = IN = 0.1 A 0
32.
From Problem 17: ZN = ZTh = 5.00  38.66
I N:
Superposition:
(E1)
ZT = Z1 + Z2  Z3
= 10  + 8  90  8  90
64  0
= 10  +
0
= very large impedance
E
Is =
=0A
ZT
and V Z1 = 0 V
with V Z2 = V Z3 = E1 = 120 V 0
120 V 0
so that Isc = E1 =
   90
8
Z3
= 15 A 90
(I)
Isc = I = 0.5 A 60
IN = Isc + Isc = + j15 A + 0.5 A 60 = + j15 A + 0.25 A + j0.433 A
= 0.25 A + j15.433 A = 15.44 A 89.07
CHAPTER 18
239
33.
a.
ZN:
E = 20 V 0, I2 = 0.4 A 20
Z1 = 6  + j8  = 10  53.13
Z2 =   j12  = 15  53.13
ZN = Z1  Z2 = (10  53.13)  (15  53.13)
= 9.66  14.93
I N:
(E)
(I2)
Isc = E/Z1 = 20 V 0/10  53.13
Isc = I2 = 0.4 A 20
= 2 A 53.13
IN = Isc + Isc = 2 A 53.13 + 0.4 A 20
= 2.15 A 42.87
34.
ZN:
E1 = 120 V 30, Z1 = 3  0
Z2 = 8   j8 , Z3 = 4  90
ZN = Z3 + Z1  Z2
= 4  90 + (3  0)  (8   j8 )
= 4.37  55.67 = 2.47  + j3.61 
I N:
I=
E1
120 V 30
=
ZT
Z1 +Z 2  Z3
120 V 30
3  + (8   j8 )  4  90
120 V 30
=
6.65  46.22
= 18.05 A 16.22
=
(I )
(8   j8 )(18.05 A   16.22)
= 22.83 A 34.65
Isc = IN = Z 2
=
8   j8  + j 4 
Z 2 + Z3
240
CHAPTER 18
35.
a.
Z1 = 212.77    90
Z3 = 22  + j47 
= 51.89   64
AC:
I N:
E
20 V 60
= 94 mA  150

Z1 212.77    90
ZN = ZTh (problem 19) = 66.04 57.36 = 35.62  + j55.61 
IN =
DC:
I N:
IN =
5V
= 227.27 mA
22 
RN = RTh = (problem 19) = 22 
b.
AC:
I N:
Z N (I N )
(66.04  57.36)(94 mA 150)

35.62   j 55.61   47 
Z N  47 
6.21 A 207.36
=
= 62.68 mA 173.22
99.08 34.14
I=
DC:
I=
3
and i = 72.46 mA + 62.68  10
Same as Problem 4 and 19.
CHAPTER 18
22 (227.27 mA)
= 72.46 mA
22   47 
sin (1000t + 173.22)
241
36.
a.
From #20
ZN = ZTh = 9  0
DC:
IN =
E
=
RT
12 V
= 1.33 A
9
AC:
IN =
=
R1 I
(6  0)(4 A 0)
=
R1 + R2
9  0
24 V 0
= 2.67 A 0
9   0
IN = 1.33 A + 2.67 A 0
b.
DC: VC = IR
= (1.33 A)(9 Ω)
= 12 V
AC: Z = 9  0  1  90
= 0.994  83.66
VC = IZ = (2.667 A 0)(0.994  83.66)
= 2.65 V 83.66
VC = 12 V + 2.65 V 83.66
37.
a.
Note Problem 21(a):
ZN = ZTh = 4.47 k 26.57
Using the same source conversion: E1 = 50 V 0
Defining ET = E1 + E = 50 V 0 + 20 V 0 = 70 V 0
Z1 = 10 k 0
Z2 = 5 k  j5 k = 7.071 k 45
Isc =
ET
70 V 0
= 7 mA 0
=
Z1 10 k  0
IN = Isc = 7 mA 0
242
CHAPTER 18
b.
( ) (4.472 k    26.565)(7 mA 0)
I = ZN IN =
Z N + Z L 4.472 k    26.565 + 5 k  90
31.30 mA   26.565 31.30 mA   26.565
=
=
4  j 2 + j5
4 + j3
31.30 mA   26.565
=
= 6.26 mA 63.44 as obtained in Problem 21.
5 36.87
38.
ZN:
Z1 = 10 k 0, Z2 = 10 k 0
Z3 = j1 k
ZN = Z3 + Z1  Z2 = 5 k  j1 k
= 5.1 k 11.31
I N:
V2 =
(Z 2  Z 3 )20 V
(Z 2  Z 3 ) + Z1
(0.995 k    84.29)(20 V)
0.1 k   j 0.99 k  + 10 k 
V2 = 1.961 V 78.69
=
IN = Isc =
39.
ZN:
V2 1.961 V   78.69
= 1.96  103 V 11.31
=
Z3
1 k    90
Z1 = 40 k 0, Z2 = 0.2 k 90
Z3 = 5 k 0
ZN = Z3  (Z1 + Z2)
= 5 k 0  (40 k  j0.2 k)
= 4.44 k 0.03
I N:
Z1 (100 I )
Z1 + Z 2
( 40 k 0)(100 I )
=
40 k   0.286
= 100 I 0.29
IN = Isc =
CHAPTER 18
243
40.
ZN:
Z1 = 5 k 0, Z2 = 1 k 90
 ZN = Z1 + Z2 = 5 k  j1 k
= 5.1 k 11.31
.
IN:
V
(20)(2 V 0)
Z1 + Z 2 5.1 k   11.31
= 7.843 mA 11.31
Isc =
(I):
=
Z1 (I )
Z1 + Z 2
(5 k 0)(2 mA 0)
=
5.1 k   11.31
= 1.96 mA 11.31
Isc =
IN = Isc + Isc = 7.843 mA 11.31 + 1.96 mA 11.31
= 9.81 mA 11.31
41.
ZN:
Z1 = 20 k 0, Z2 = 5 k 0
V = 10 V 0, μ = 20, h = 100
I = 1 mA 0
ZN = Z1 + Z2 = 25 k 0
IN: (hI)
Z1 (hI )
Z1 + Z 2
(20 k 0)(hI )
=
20 k 0 + 5 k0
= 80 mA 0
Isc =
(μV)
Isc =
V
=
Z1 + Z 2
= 8 mA 0
(20)(10 V 0)
25 k
IN (direction of Isc) = Isc  Isc = 80 mA 0  8 mA 0 = 72 mA 0
244
CHAPTER 18
42.
Z1 = 2 k 0
Z2 = 5 k 0
I2 = I3 + I5
V = I5Z2 = (I2  I3)Z2
Eoc = ETh = 21 V = 21(I2  I3)Z2

E 
= 21 I 2  oc  Z 2
Z1 


Z2 
Eoc 1 + 21  = 21 Z2I2
Z1 

21 Z 2 I 2 21(5 k  0)(2 mA 0)
=
Eoc =
Z
 5 k 0 
2
1 + 21
1 + 21

Z1
 2 k 0 
ETh = Eoc = 3.925 V 0
20 V  V  V = 0
and IN = Isc = I2 = 2 mA 0
3.925 V 0
ZN = Eoc =
= 1.96 k
2 mA 0
I sc
43.
Z1 = 1 k 0
Z2 = 3 k 0
Z3 = 4 k 0
Eoc
21
Eoc
V
=
I = I1 + I2, I1 =
Z1
21 Z1
V2 = 21 V = Eoc  V =
CHAPTER 18
245
I2 =
and
 1
Eoc
Eoc
E
1 
, I = I1 + I2 =
+ oc = Eoc 
+

Z2
21 Z1
Z2
 21 Z1 Z 2 
 + 21 Z1 
I = Eoc  Z 2

 21 Z1Z 2 
21 Z1Z 2 I
(21)(1 k  0)(3 k  0)(2 mA 0)
Eoc =
=
Z 2 + 21Z1
3 k  + 21(1 k  0)
ETh = Eoc = 5.25 V 0
Z
V3 21 V
 V = 3 Isc
=
Z3
Z3
21
V = I1Z1
I = I1 + I
Isc =
Isc =
 Z + Z3 
Z 2 I
 I =  2
 I sc
Z 2 + Z3
 Z2 
I = I1 + I =
Isc =

 Z3
Z + Z3 
V  Z 2 + Z3 
+ 2
+
 I sc = 
 I sc
Z1  Z 2 
Z2 
 21 Z1
I
2 mA 0
= 0.79 mA 0
=
Z3
Z3 + Z 2
4 k 7 k
+
+
21 k  3 k 
Z2
21 Z1
IN = 0.79 mA 0
E
5.25 V 0
= 6.65 k 0
ZN = oc =
I sc 0.79 mA 0
44.
Z1 = 3  + j4 , Z2 = j6 
 ZTh = Z1  Z2
= 5  53.13  6  90
= 8.32  3.18
ZL = 8.32  3.18 = 8.31   j0.46 
246
CHAPTER 18
ETh =
Z2 E
Z 2 + Z1
(6    90)(120 V 0)
3.61    33.69
= 199.45 V 56.31
2
(3.124 V ) 2
= 1198.2 W
Pmax = E Th =
4RTh
4(8.31 )
=
Z1 = 3  + j4  = 5  53.13
Z2 = 2 Ω 0
 ZN = ZTh = Z1  Z2
= 5  53.13  2  0
10  53.13
=
2 + 3 + j4
10 Ω 53.13
=
5 + j4
10  53.13
=
6.403 38.66
= 1.56  14.47
ZTh = 1.56  14.47
= 1.51  + j0.39 
ZL = 1.51   j0.39 
45.
ETh = I(Z1 Z2)
= (2 A 30)(1.562  14.47)
= 3.12 V 44.47
2
(3.12 V ) 2
= 1.61 W
Pmax = E Th =
4RTh 4(1.51 )
46.
ZTh:
Z1 = 4  90, Z2 = 10  0
Z3 = 5  90, Z4 = 6  90
E = 60 V 60
ZTh = Z4 + Z3  (Z1 + Z2) = j6  + (5  90)  (10  + j4 )
= 2.475   j4.754 
= 11.04  77.03
ZL = 11.04  77.03
CHAPTER 18
247
ETh:
ETh =
Z 3 (E )
Z3 + Z1 + Z 2
( 5    90)(60 V 60)
 j5  + j 4  + 10 
= 29.85 V 24.29
=
2
/ 4 RTh = (29.85 V)2/4(2.475 ) = 90 W
Pmax = E Th
47.
Z1 = 3  + j4  = 5  53.13
Z2 = j8 
Z3 = 12  + j9 
ZTh = Z2 + Z1  Z3 = j8  + (5  53.13)  (15  36.87)
= 5.71  64.30 = 2.475   j5.143 
ZL = 5.71  64.30 = 2.48 Ω + j5.15 
ETh + V Z3  E2 = 0
ETh = E2  VZ3
Z3 (E 2  E1 )
Z3 + Z1
= 168.97 V 112.53
VZ3 =
ETh = E2  VZ3 = 200 V 90  168.97 V 112.53 = 78.24 V 34.16
2
/ 4 RTh = (78.24 V)2/4(2.475 ) = 618.33 W
Pmax = E Th
48.
248
E  0 1 V  0
= 1 mA 0
=
R10 1 k  0
ZTh = 40 k 0
ETh = (50 I)(40 k 0) = (50)(1 mA 0)(40 k 0) = 2000 V 0
2
( 2 kV ) 2
= 25 W
Pmax = E Th =
4 RTh 4(40 k )
I=
CHAPTER 18
49.
ETh:
Z1 = 2 k 0
Z2 = = 3 k 90
Z3 = 6 k 90
Z2E
(3 k 90)(20 V 0)

Z 2  Z1
 j 3 k  2 k
60 V   90
= 16.62 V 33.69

3.61   56.31
ETh 
ZTh:
 ZTh  Z3  Z1  Z 2
(2 k 0)(3 k   90)
2 k  j 3 k
  j 6 k  1.66 k   33.69
ZTh = +j6 k +
  j 6 k  1.38 k  j 920.8 
 1.38 k  j 5.08 k
 5.26 k   74.80
ZL = 5.36 k 74.80 = 1.38k  j5.08 k
b.
50.
Pmax =
2
ETh
(16.62 V) 2
= 50.04 mW

4 RTh 4(1.38 k)
From #20, ZTh = 9 , ETh = 12 V + 24 V 0
a.
 ZL = 9 
b.
2
(12 V ) 2 (24 V ) 2
Pmax = E Th =
= 4 W + 16 W = 20 W
+
4 RTh 4(9 )
4(9 )
or ETh = V 02 + V12eff
2
= 26.833 V
2
(26.833 V )
and Pmax = E Th =
= 20 W
4 RTh
4(9 )
CHAPTER 18
249
51.
52.
a.
Problem 21(a):
ZTh = 4.47 k 26.57 = 4 k  j2 k
ZL = 4 k + j2 k
ETh = 31.31 V 26.57
b.
2
/ 4 RTh = (31.31 V)2/4(4 k) = 61.27 mW
Pmax = E Th
a.
ZTh = 2 k 0  2 k 90 = 1 k  j1 k
R L = R Th +  X Th + X Load 
2
2
= (1 k) 2 + (1 k + 2 k) 2
= (1 k) 2 + (1 k) 2
= 1.41 k
53.
b.
Rav = (RTh + RLoad)/2 = (1 k + 1.41 k)/2 = 1.21 k
2
(50 V ) 2
= 516.53 mW
Pmax = E Th =
4 Rav 4(1.21 k)
a.
ZTh:
1
1
=
2 fC 2 (10 kHz)(4 nF)
 3978.87 
XL = 2fL = 2(10 kHz)(30 mH)
 1884.96 
Z1 = 1 k 0, Z2 = 1884.96  90
Z3 = 3978.87  90
ZTh = (Z1 + Z2)  Z3 = (1 k + j1884.96 )  3978.87  −90)
= 2133.79  62.05  3978.87  −90)
= 3658.65  36.52
XC =
 ZL = 3658.65  36.52 = 2940.27  − j2177.27 
1
1
= 7.31 nF
C=
=
2 fX C 2 (10 kHz)(2177.27 )
b.
RL = RTh = 2940.27 
c.
ETh 
Z3 (E)
(3978.87    90)(2 V0)
=
= 3.43 V−25.53)
Z3  Z1  Z 2 1 k + j1884.96   j 3978.87
2
/ 4 RTh = (3.43 V) 2 /4(2940.27 Ω) = 1 mW
Pmax = ETh
250
CHAPTER 18
(4 k 0)(4 mA 0)
= 1.33 mA 0
4 k + 8 k
Vab = (Iab)(8 k 0) = 10.67 V 0
54.
Iab =
55.
a.
4 k(E)
1
= (20 V 0)
4 k + 12 k 4
= 5 V 0
5 V 0
= 0.83 mA 0
I=
6 k
V=
b.
6 k ( E )
1
= (20 V 0)
2
6 k + 6 k
= 10 V 0
10 V 0
= 0.83 mA 0
I=
12 k
V=
56.
100 V 0
= 50 mA 0
2 k   0
50 V 0
I2 =
4 k  90
= 12.5 mA 90
Z1 = 2 k 0
Z2 = 4 k 90
Z3 = 4 k 90
IT = I1  I2 = (50 mA 0  12.5 mA 90) = 50 mA + j12.5 mA
= 51.54 mA 14.04
Z = Z1  Z2 = (2 k 0)  (4 k 90) = 1.79 k 26.57
ZIT
(1.79 k  26.57)(51.54 mA 14.04)
IC =
=
1.6 k + j 0.8 k  j 4 k
Z + Z 3
= 25.77 mA 104.04
I1 =
CHAPTER 18
251
Chapter 19
1.
a.
PT = 60 W + 45 W + 25 W = 130 W
b.
QT = 0 VARS, ST = PT = 130 VA
c.
130 VA
S
= 0.542 A
ST = EIs, Is = T =
E
240 V
d.
60 W
= 204.2 
(0.542 A ) 2
V = IsR = (0.542 A)(204.2 ) = 110.68 V
V1 = V2 = E  V = 240 V  110.68 V = 129.32 V
2
2
(129.32 V ) 2
V
P1 = V 1 , R1 = 1 =
= 371.6 
45 W
R1
P1
P = I s2 R, R =
P
2
Is
=
2
e.
2.
a.
2
2
(129.32 V )
V
P2 = V 2 , R2 = 2 =
= 668.9 
25 W
R2
P2
V 129.32 V
V 129.32 V
= 0.348 A, I2 = 2 =
= 0.193 A
I1 = 1 =
R1 371.6 
R2 668.9 
ZT = 3   j5  + j9  = 3  + j4  = 5  53.13
E
50 V 0
= 10 A 53.13
I=
=
ZT 5  53.13
R:
L:
C:
P = I2R = (10 A)2 3  = 300 W
P=0W
P=0W
b.
R:
C:
L:
Q = 0 VAR
QC = I2XC = (10 A)2 5  = 500 VAR
QL = I2XL = (10 A)2 9  = 900 VAR
c.
R:
C:
L:
S = 300 VA
S = 500 VA
S = 900 VA
d.
PT = 300 W
QT = QL  QC = 400 VAR(L)
ST =
2
2
PT + QT = EI = (50 V)(10 A) = 500 VA
300 W
= 0.6 lagging
Fp = PT =
S T 500 VA
e.
252

CHAPTER 19
f.
WR =
 VI 
 VI  VI
VI
: WR = 2   = 2 
=
f1
 f2 
 2 f1  f1
V = IR = (10 A)(3 ) = 30 V
(30 V)(10 A)
=5J
WR =
60 Hz
3.
g.
VC = IXC = (10 A)(5 ) = 50 V
VI (50 V)(10 A)
= 1.33 J
WC =
=
1 (2 )(60 Hz)
VL = IXL = (10 A)(9 ) = 90 V
VI (90 V)(10 A)
WL =
= 2.39 J
=
1
376.8
a.
PT = 0 + 100 W + 300 W = 400 W
QT = 200 VAR(L)  600 VAR(C) + 0 = 400 VAR(C)
b.
ST =
PT2 + QT2 = 565.69 VA
Fp =
PT
400 W
= 0.707 (leading)
=
ST 565.69 VA

PT = EIs cos θT
400 W = (100 V)Is(0.7071)
400 W
= 5.66 A
Is =
70.71 V
Is = 5.66 A 135
c.
4.
5.
a.
PT = 600 W + 500 W + 100 W = 1200 W
QT = 1200 VAR(L) + 600 VAR(L)  1800(C) = 0 VAR
ST = PT = 1200 VA
b.
1200 W
Fp = P T =
=1
S T 1200 VA
c.

d.
Is =
a.
PT = 200 W + 100 W + 0 + 50 W = 350 W
QT = 50 VAR(L) + 100 VAR(L)  200 VAR(C)  400 VAR(C) = 450 VAR(C)
ST =
CHAPTER 19
S T 1200 VA
= 6 A, 1  0
=
E
200 V
Is = 6 A 0
2
2
PT + QT = 570.09 VA
253
6.
b.
350 W
Fp = P T =
= 0.614 (leading)
S T 570.09 VA
c.

d.
PT = EIs cos θT
350 W = (50 V)Is(0.614)
350 W
= 11.4 A
Is =
30.7 V
Is = 11.4 A 52.12
a.
IR =
b.
c.
60 V 30
= 3 A 30
20  0
P = I2R = (3 A)2 20  = 180 W
QR = 0 VAR
S = P = 180 VA
60 V 30
= 6 A 60
10  90
PL = 0 W
QL = I2XL = (6 A)2 10  = 360 VAR(L)
S = Q = 360 VA
IL =
PT = 180 W + 400 W = 580 W
QT = 600 VAR(L) + 360 VAR(L) = 960 VAR(L)
ST =
(580 W) 2 + (960 VAR) 2 = 1121.61 VA
580 W
Fp = PT =
= 0.517 (lagging) θ = 58.87
1121.61
VA
ST
d.
ST = EIs
1121.61 VA
S
= 18.69 A
Is = T =
E
60 V
 I s = 30  58.87 = 28.87
Is = 18.69 A 28.87
7.
a.
b.
254
2
(20 V)2
R: P = E =
= 200 W
R
2
PL,C = 0 W
R:
Q = 0 VAR
C:
QC = E =
XC
L:
QL = E =
XL
2
(20 V)2
= 80 VAR(C)
5
2
(20 V ) 2
= 100 VAR(L)
4
CHAPTER 19
c.
R:
C:
L:
d.
PT = 200 W + 0 + 0 = 200 W
QT = 0 + 80 VAR(C) + 100 VAR(L) = 20 VAR(L)
ST =
S = 200 VA
S = 80 VA
S = 100 VA
(200 W)2 + (20 VAR) 2 = 200 VA
200 W
Fp = PT =
= 0.995 (lagging)  5.73
S T 200.998 VA
8.
e.

f.
Is =
a.
S T 200.998 VA
=
= 10.05 A
E
20 V
Is = 10.05 A5.73
50 V 60
= 10 A 6.87
5  53.13
PR = I2R = (10 A)2 3  = 300 W
PL = 0 W
PC = 0 W
R  L:
I=
b.
QR = 0 VAR
QL = I2XL = (10 A)2 4  = 400 VAR
50 V 60
= 5 A 150
IC =
10    90
QC = I2XC = (5 A)2 10 Ω = 250 VAR
c.
SR = P = 300 VA
SL = QL = 400 VA
SC = QC = 250 VA
d.
PT = PR = 300 W
QT = 400 VAR(L)  250 VAR(C) = 150 VAR(L)
ST =
(300 W) 2 + (150 VAR) 2 = 335.41 VA
300 W
Fp = PT =
= 0.894 (lagging)
S T 335.41 VA
e.
f.

S T 335.41 VA
=
= 6.71 A
E
50 V
0.894  26.62 lagging
θ = 60  26.62 = 33.38
Is = 6.71 A 33.38
Is =
CHAPTER 19
255
9. ac.
XL = ωL = (400 rad/s)(0.1 H) = 40 
1
1
XC =
=
C (400 rad/s)(100  F)
= 25 
Z1 = 40  90, Z2 = 25  90
Z3 = 30  0
ZT = Z1 + Z2  Z3 = +j40  + (25  90)  (30  0)
= +j40  + 19.21  50.19
= +j40  + 12.3   j14.76 
= 12.3  + j25.24 
= 28.08  64.02
Is =
E
50 V 0
= 1.78 A 64.02
28.08  64.02
=
ZT
V2 = Is(Z2  Z3) = (1.78 A 64.02)(19.21  50.19)
= 34.19 V 114.21
34.19 V   114.21
I2 = V 2 =
= 1.37 A 24.21
25    90
Z2
34.19 V   114.21
= 1.14 A 114.21
I3 = V 2 =
30  0
Z3
d.
Z1:
P = 0 W, QL = I s2 X L = (1.78 A)2 40  = 126.74 VAR(L), S = 126.74 VA
Z2:
P = 0 W, QC = I 22 X C = (1.37 A)2 25  = 46.92 VAR(C), S = 46.92 VA
Z3:
P = I 32 R = (1.14 A)2 30  = 38.99 W, QR = 0 VAR, S = 38.99 VA
PT = 0 + 0 + 38.99 W = 38.99 W
QT = +126.74 VAR(L)  46.92 VAR(C) + 0 = 79.82 VAR(L)
2
2
ST =
PT + QT = 88.83 VA
38.99 W
= 0.439 (lagging)
Fp = PT =
S T 88.83 VA
e.

f.
WR =
f1 =
g.
1 400 rad/s
= 63.69 Hz
=
2
6.28
WL =
WC =
256
VR I R V2 I 3 (34.19 V)(1.14 A)
= 0.31 J
=
=
2f1
2 f1
2(63.69 Hz)
VL I L
=
(Is X L )Is
VC I C
=
V2 I 2
1
1
1
1
=
=
I s2 X L
1
=
(1.78 A) 2 40 
= 0.32 J
400 rad/s
(34.19 V)(1.37 A)
= 0.12 J
400 rad/s
CHAPTER 19
10.
a.
b.
11.
12.
a.
10,000 VA
Is = S T =
= 50 A
E
200 V
0.5  60 leading
 Is leads E by 60
E
200 V 0
= 4 Ω 60 = 2   j3.464  = R  jXC
ZT =
=
50 A 60
Is
Fp = PT  PT = FpST = (0.5)(10,000 VA) = 5000 W
ST
S T 5000 VA
=
= 41.67 A
E
120 V
Fp = 0.8  36.87 (lagging)
E = 120 V 0, I = 41.67 A 36.87
E
120 V 0
= 2.88  36.87 = 2.30  + j1.73  = R + jXL
Z= =
I 41.67 A   36.87
I=
b.
P = S cos θ = (5000 VA)(0.8) = 4000 W
a.
PT = 0 + 300 W = 300 W
QT = 600 VAR(C) + 200(L) = 400 VAR(C)
ST =
2
2
PT + QT = 500 VA
P
300 W
Fp = T =
= 0.6 (leading)
ST 500 VA
b.
ST 500 VA
=
= 16.67 A
E
30 V
Fp = 0.6  53.13
Is = 16.67 A 53.13
Is =
c.

d.
Load: 600 VAR(C), 0 W
R = 0, L = 0, QC = I2XC  XC =
QC
2
=
600 VAR
= 2.159 
(16.67 A)2
I
Load: 200 VAR(L), 300 W
C = 0, R = P/I2 = 300 W/(16.67 A)2 = 1.079 
Q
200 VAR
= 0.7197 
XL = 2L =
(16.67 A) 2
I
ZT = j2.159  + 1.0796  + j0.7197 
= 1.08   j1.44 
CHAPTER 19
257
13.
a.
PT = 0 + 300 W + 600 W = 900 W
QT = 500 VAR(C) + 0 + 500 VAR(L) = 0 VAR
ST = PT = 900 VA
Fp = PT = 1
ST
b.
Is =
c.

S T 900 VA
=
= 9 A, Is = 9 A 0
E
100 V
d.
2
Z1:
Z3:
14.
a.
4
100 V 0
= 5A 90
Z1 20    90
I2 = Is  I1 = 9 A  j5 A = 10.296 A 29.05
P
300 W
300
R = 2=
= 2.83 
=
2
106
(10.296
A)
I
XL,C = 0 
P
600 W
R= 2=
= 5.66 
I 2 (10.296 A)2
Q
500
XL = 2 =
= 4.72 , XC = 0 
I 2 (10.296 A)2
I1 =
Z2:
2
QC = V  X C = V = 10 = 20 
XC
QC 500
E
=
PT = 200 W + 30 W + 0 = 230 W
QT = 0 + 40 VAR(L) + 100 VAR(L) = 140 VAR(L)
ST =
2
2
PT + QT = 269.26 VA
230 W
= 0.854 (lagging)  31.35
Fp = PT =
S T 269.26 VA
b.
258
S T 269.26 VA
= 2.6926 A
=
E
100 V
Is = 2.69 A 31.35
Is =
CHAPTER 19
c.
15.
a.
2
4
R = V = 10 = 50 
P 200
XL,XC = 0 
100 V 0
= 2 A 0
I1 =
50  0
I2 = Is  I1
= 2.6926 A 31.35  2 A 0
= 2.299 A  j1.40 A  2.0 A
= 0.299 A  j1.40 A
= 1.432 A 77.94
Z1:
P
30 W
Q
40 VAR
= 14.63 , XL = 2 =
= 19.50 
=
2
2
I 2 (1.432 A)
I 2 (1.432 A)2
XC = 0 
Z2:
R=
Z3:
XL =
Q 100 VAR
= 48.76 , R = 0 , XC = 0 
=
I 22 (1.432 A)2
PT = 100 W + 1000 W = 1100 W
QT = 75 VAR(C) + 2291.26 VAR(C) = 2366.26 VAR(C)
ST =
2
2
PT + QT = 2609.44 VA
1100 W
= 0.422 (leading)  65.04
Fp = PT =
S T 2609.44 VA
b.
S T 2609.44 VA
= 521.89 V
=
I
5A
E = 521.89 V 65.07
ST = EI  E =
c.
I Z1 =
S S 125 VA
= 0.2395 A
= =
V1 E 521.89 V
I Z2 =
S S 2500 VA
= 4.79 A
= =
V2 E 521.89 V
CHAPTER 19
259
Z1:
R=
P
100 W
= 1743.38 
=
2
I Z1 (0.2395)2
Q = I Z21 X C  X C =
Z2:
R=
XC =
16.
P
I Z21 X C
=
Q
I Z21 X C
Q
75 VAR
= 1307.53 
=
2
I Z1 (0.2395 A)2
1000 W
= 43.59 
(4.790 A) 2
=
2291.26 VAR
= 99.88 
(4.790 A) 2
a.
0.7  45.573
P = S cos θ = (10 kVA)(0.7) = 7 kW
Q = S sin θ = (10 kVA)(0.714) = 7.14 kVAR(L)
b.
QC = 7.14 kVAR =
V2
XC
V2
(208 V) 2
= 6.059 
=
Q C 7.14 kVAR
1
1
1
XC =
 C=
= 438 μF
=
2 fC
2 fX C (2π)(60 Hz)(6.059 )
XC =
c.
Uncompensated:
10,000 VA
S
= 48.08 A
Is = T =
E
208 V
Compensated:
7,000 W
S
= 33.65 A
Is = T = P T =
E
E
208 V
d.
260
cos θ = 0.9
θ = cos10.9 = 25.842
x
tan θ =
7 kW
x = (7 kW)(tan 25.842)
= (7 kW)(0.484)
= 3.39 kVAR
y = (7.14  3.39) kVAR
= 3.75 kVAR
CHAPTER 19
QC = 3.75 kVAR =
V2
XC
XC =
V2
(208 V)2
= 11.537 
=
QC 3.75 kVAR
C=
1
1
= 230 μF
=
2 fX C (2π)(60 Hz)(11.537 )
Uncompensated:
Is = 48.08 A
Compensated:
ST =
(7 kW) 2 + (3.39 kVAR) 2 = 7.778 kVA
S T 7.778 kVA
=
= 37.39 A
E
208 V
Is = 48.08 A  37.39 A = 10.69 A
Is =
17.
a.
PT = 5 kW, QT = 6 kVAR(L)
ST =
2
2
PT + QT = 7.81 kVA
b.
5 kW
= 0.640 (lagging)
Fp = PT =
S T 7.81 kVA
c.
7,810 VA
Is = S T =
= 65.08 A
E
120 V
d.
XC =
2
1
(120 V) 2
, QC = I2XC = E =
2 fC
XC
XC
and
e.
(120 V)2 14, 400
= 2.4 
=
6000
QC
1
1
= 1105 μF
C=
=
2 fX C (2 )(60 Hz)(2.4 )
XC =
ST = EIs = PT
5000 W
 Is = P T =
= 41.67 A
E
120 V
CHAPTER 19
261
18.
a.
Load 1:
Load 2:
P = 20,000 W, Q = 0 VAR
θ = cos10.7 = 45.573
x
10 kW
x = (10 kW)tan 45.573
= (10 kW)(1.02)
= 10,202 VAR(L)
tan θ =
θ = cos10.85 = 31.788
Load 3:
x
5 kW
x = (5 kW)tan 31.788
= (5 kW)(0.62)
= 3098.7 VAR(L)
tan θ =
PT = 20,000 W + 10,000 W + 5,000 W = 35 kW
QT = 0 + 10,202 VAR + 3098.7 VAR = 13,300.7 VAR(L)
ST =
b.
2
QC = QL = 13,300.7 VAR
2
(103 V)2
XC = E =
= 75.184 
Q C 13,300.7 VAR
C=
c.
2
PT + QT = 37,442 VA = 37.442 kVA
1
1
= 35.28 μF
=
2 fX C (2π)(60 Hz)(75.184 )
Uncompensated:
37.442 kVA
Is = S T =
= 37.44 A
E
1 kV
Compensated:
ST = PT = 35 kW
35 kW
S
= 35 A
Is = T =
E
1 kV
⌬Is = 37.44 A  35 A = 2.44 A
19.
262
a.
ZT = R1 + R2 + R3 + jXL  jXC
= 2  + 3  + 1  + j3   j12  = 6   j9  = 10.82  56.31
E
50 V 0
=
= 4.62 A 56.31
I=
ZT 10.82    56.31
CHAPTER 19
b.
20.
21.
P = VI cos θ = (50 V)(4.62 A) cos 56.31 = 128.14 W
a-b: P = I2R = (4.62 A)2 2  = 42.69 W
b-c: P = I2R = (4.62 A)2 3  = 64.03 W
a-c: 42.69 W + 64.03 W = 106.72 W
a-d: 106.72 W
c-d: 0 W
d-e: 0 W
f-e: P = I2R = (4.62 A)2 1  = 21.34 W
a.
ST = 660 VA = EIs
660 VA
= 5.5 A
Is =
120 V
θ = cos10.6 = 53.13
 E = 120 V 0, Is = 5.5 A 53.13
P = EI cos θ = (120 V)(5.5 A)(0.6) = 396 W
Wattmeter = 396 W, Ammeter = 5.5 A, Voltmeter = 120 V
b.
ZT =
a.
R=
E
120 V 0
= 21.82  53.13 = 13.09  + j17.46  = R + jXL
=
I
5.5 A   53.13
P
I
2
b.
R=
XL
49.75 
= 132.03 mH
=
2 f (2π)(60 Hz)
P
I
c.
R=
2
P
I
2
22.
a.
=
90 W
= 10 
(3 A) 2
=
E 200 V
60 W
= 15 , ZT = =
= 100 
2
I
2A
(2 A)
ZT2  R 2 = (100 Ω) 2  (15 Ω)2 = 98.87 
XL =
L=
E 200 V
80 W
= 5 Ω, ZT = =
= 50 
2
I
4A
(4 A )
ZT2  R 2 = (50 ) 2  (5 Ω) 2 = 49.75 
XL =
L=
=
XL
98.87 
= 262.39 mH
=
2 f
376.8
XL = 2πfL = (6.28)(50 Hz)(0.08 H) = 25.12 
R 2 + X L2 = (4 Ω) 2 + (25.12 )2 = 25.44 
ZT =
60 V
= 2.358 A
Z T 25.44 
P = I2R = (2.358 A)2 4  = 22.24 W
I=
CHAPTER 19
E
=
263
b.
P
30 W
= 2.07 A
=
7
R
E
60 V
= 28.99 
ZT = =
I 2.07 A
I=
XL =
L=
c.
XL
28.13 Ω
=
= 89.54 mH
2 f (2π)(50 Hz)
P = I2R = (1.7 A)2 10  = 28.9 W
E 60 V
= 35.29 
ZT = =
I 1.7 A
XL =
L=
264
(28.99 Ω) 2  (7 Ω) 2 = 28.13 
(35.29 Ω)2  (10 Ω) 2 = 33.84 
XL
38.84 
= 107.77 mH
=
2 f
314
CHAPTER 19
Chapter 20
1.
a.
ωs =
s 250 rad/s
= 39.79 Hz
=
2
2
fs =
b.
ωs =
3.
1
= 3496.50 rad/s
(0.51 H)(0.16  F)
s 3496.50 rad/s
= 556.49 Hz
=
2
2
fs =
2.
1
1
=
= 250 rad/s
LC
1 H)(16  F)
1
= 22,173 rad/s
(0.27 mH)(7.5  F)
 22,173 rad/s
= 3528.93 Hz
fs = s =
2
2
c.
ωs =
a.
XC = 30 
d.
VR = IR = (25 mA)(10 ) = 250 mV = E
VL = IXL = (25 mA)(30 ) = 750 mV
VC = IXC = (25 mA)(30 ) = 750 mV
VL = VC
e.
Qs =
a.
XL = 2 k
b.
I=
c.
VR = IR = (120 mA)(100 ) = 12 V = E
VL = IXL = (120 mA)(2 k) = 240 V
VC = IXC = (120 mA)(2 k) = 240 V
VL = VC = 20 VR
d.
Qs =
e.
b.
ZTs = 2 
X L 30 
=
= 15 (med Q)
R
2
c.
f.
I=
E 50 mV
= 25 mA
=
2
ZTs
P = I2R = (25 mA)2 2  = 1.25 mW
E
12 V
= 120 mA
=
ZTs 100 
X L 20000 
=
= 20 (high Q)
R
100 
XL
2 k
= 63.7 mH
=
2 f 2 (5 kHz)
1
1
1
,C=
=
= 15,920 pF
XC =
2 fC
2 fX C 2 (5 kHz)(2 k)
XL = 2πfL, L =
CHAPTER 20
265
f.
g.
4.
f s 5 kHz
= 250 Hz
=
20
Qs
BW
0.25 kHz
= 5 kHz +
= 5.13 kHz
2
2
BW
0.25 kHz
= 5 kHz 
= 4.88 kHz
f1 = fs 
2
2
f2 = fs +
1
 L=
1
1
= 3.91 mH
=
2
(2 f s ) C (2 1.8 kHz ) 2 2  F
a.
fs =
b.
XL = 2πfL = 2π(1.8 kHz)(3.91 mH) = 44.2 
1
1
=
= 44.2 
XC =
2 fC 2 (1.8 kHz)(2  F)
XL = XC
c.
Erms = (0.707)(20 mV) = 14.14 mV
E
14.14 mV
= 3.01 mA
Irms = rms =
R
4.7 
d.
P = I2R = (3.01 mA)2 4.7  = 42.58 μW
e.
ST = PT = 42.58 μVA
g.
h.
266
BW =
2 LC
f.
Fp = 1
44.2 
Qs = X L =
= 9.4
R
4.7 
f
1.8 kHz
= 191.49 Hz
BW = s =
9.4
Qs
2
1  R 1 R
4 
 +

+
 
2  2 L 2  L  LC 


2

1  4.7 
1  4.7  
4


+
+
=


2  2(3.91 mH) 2  3.91 mH  (3.91 mH)(2  F) 


1
601.02  11.324  103 
=

2 
= 1897.93 Hz
2
1  R 1 R
4 


f1 =
+
+
 
2  2 L 2  L  LC 


1
 601.02  11.324  103 
=

2 
= 1.71 kHz
1
1
PHPF = Pmax = (42.58 μW) = 21.29 μW
2
2
f2 =
CHAPTER 20
5.
a.
b.
6.
BW = fs/Qs = 6000 Hz/15 = 400 Hz
BW
= 6000 Hz + 200 Hz = 6200 Hz
2
BW
f1 = fs 
= 6000 Hz  200 Hz = 5800 Hz
2
f2 = fs +
XL
 XL = QsR = (15)(3 ) = 45  = XC
R
c.
Qs =
d.
PHPF =
a.
L=
1
1
1
Pmax = (I2R) = (0.5 A)2 3 = 375 mW
2
2
2
XL
200 
= 3.185 mH
=
2 f 2 (104 Hz)
5
 250 Hz
2 L 2 (3.185 mH)
200 
f
10,000 Hz
= 40, BW = s =
= 250 Hz
or Qs = X L = X C =
R
R
5
40
Qs
BW =
b.
7.
R
=
f2 = fs + BW/2 = 10,000 Hz + 250 Hz/2 = 10,125 Hz
f1 = fs  BW/2 = 10,000 Hz  125 Hz = 9,875 Hz
c.
200 
= 40
Qs = X L =
R
5
d.
I=
e.
P = I2R = (6 A)2 5  = 180 W
a.
BW =
b.
Qs = X L  XL = QsR = (10)(2 ) = 20 
R
c.
L=
E 0 30 V 0
= 6 A 0, VL = (I 0)(XL 90)
=
R0 5  0
= (6 A 0)(200  90)
= 1200 V 90
VC = (I 0)(XC 90) = 1200 V 90
fs
 Qs = fs/BW = 2000 Hz/200 Hz = 10
Qs
XL
20 
=
= 1.59 mH
2 f (6.28)(2 kHz)
1
1
= 3.98 μF
C=
=
2 fX C (6.28)(2 kHz)(20 )
CHAPTER 20
267
8.
9.
d.
f2 = fs + BW/2 = 2000 Hz + 100 Hz = 2100 Hz
f1 = fs  BW/2 = 2000 Hz  100 Hz = 1900 Hz
a.
BW = 6000 Hz  5400 Hz = 600 Hz
b.
BW = fs/Qs  fs = QsBW = (9.5)(600 Hz) = 5700 Hz
c.
Qs =
XL
 XL = XC = QsR = (9.5)(2 ) = 19 
R
d.
L=
XL
19 
= 0.53 mH
=
2 f 2 (5700 Hz)
C=
1
1
= 1.47 μF
=
2 fX C 2 (5.7 kHz)(19 )
IM =
E
E
5V
R=
=
= 10 
R
500 mA
IM
BW = fs/Qs  Qs = fs/BW = 8400 Hz/120 Hz = 70
X
Qs = L  XL = QsR = (70)(10 Ω) = 700 
R
XC = XL = 700 
XL
700 
= 13.26 mH
L=
=
2 f (2π)(8.4 kHz)
1
1
C=
= 27.07 nF
=
2 fX C (2π)(8.4 kHz)(0.7 k )
f2 = fs + BW/2 = 8400 Hz + 120 Hz/2 = 8.46 kHz
f1 = fs  BW/2 = 8400 Hz  60 Hz = 8.34 kHz
10.
Qs = X L  XL = QsR = 20(2 Ω) = 40  = XC
R
f
BW = s  fs = QsBW = (20)(400 Hz) = 8 kHz
Qs
40 
= 795.77 H
L= XL =
2 f 2 (8 kHz)
1
1
C=
= 497.36 nF
=
2 fX C 2 (8 kHz)(40 )
f2 = fs + BW/2 = 8000 Hz + 400 Hz/2 = 8200 Hz
f1 = fs  BW/2 = 8000 Hz  200 Hz = 7800 Hz
11.
268
s 2  106 rad/s
=
= 1 MHz
2
2
a.
fs =
b.
f 2  f1
= 0.16  BW = f2  f1 = 0.16 fs = 0.16(1 MHz) = 160 kHz
fs
CHAPTER 20
c.
12.
2
2
(120 V)2
P= VR R= VR=
= 720 
R
P
20 W
R
R
720 
BW =
 L=
= 0.716 mH
=
2 L
2 BW (6.28)(160 kHz)
1
1
1
fs =
C=
= 2 6
= 35.38 pF
2 2
4 f s L 4 (10 Hz ) 2(0.716 mH)
2 LC
d.
X
2 f s L 2 (106 Hz)(0.716 mH)
Q = X L = 80  R = L =
= 56.23 
=
80
80
80
R
a.
Q =
XL
R
X
2 fL 2 (1MHz)(100  H)
R = L =
= 50.27 
=
12.5
Q
Q
f 2  f1 1
=
= 0.2
fs
Qs
X
1
2 fL 2 (1 MHz)(100  H) 628.32 
=
= 5= L =
=
0.2
R
R
R
R
628.32 
= 125.66
R=
5
R = Rd + R
Qs =
125.66  = Rd + 50.27 
and Rd = 125.66   50.27  = 75.39 
13.
c.
XC =
a.
fp =
1
= XL
2 fC
1
1
C=
= 253.3 pF
=
2 fX C 2 (1 MHz)(628.32 )
1
2 LC
=
2
= 159.16 kHz
2 (0.1 mH)(10 nF)
b.
c.
IL =
VL
4V
4V
=
= 40 mA
=
X L 2 f p L 100 
IC =
VL
4V
4V
=
= 40 mA
=
X C 1/ 2 f p C 100 
CHAPTER 20
269
14.
15.
d.
2 k
2 k
Qp = R s =
=
= 20
X Lp 2 f p L 100 
a.
fs =
b.
Q =
c.
Since Q  10, fp  fs = 13.4 kHz
d.
XL = 2πfpL = 2π(13.4 kHz)(4.7 mH) = 395.72 
1
1
=
= 395.91 
XC =
2 f p C 2 (13.4 kHz)(30 nF)
XL = XC
e.
ZTp = Q2 R = (49.46)2 8  = 19.57 k
f.
VC = IZ T p = (10 mA)(19.57 k) = 195.7 V
g.
Q  10, Qp = Q = 49.46
f p 13.4 kHz
BW =
= 270.9 Hz
=
49.46
Qp
h.
IL = IC = Q IT = (49.46)(10 mA) = 494.6 mA
a.
fs =
b.
Q =
1
2 LC
=
1
= 13.4 kHz
2 (4.7 mH)(30 nF)
X L 2 fL 2 (13.4 kHz)(4.7 mH)
= 49.46  10 (yes)
=
=
8
R
R
1
2 LC
=
1
= 1.027 MHz
2 (200  H)(120  F)
XL
X
2 (1.027 MHz)(200  H)
= 86.04 
 R  L 
15
R
Q
Zp= Q2 R  (15) 2 86.04  = 19.36 k
c.
270
P = I R = (120 mA) (950.9 ) = 13.69 W
2
2
CHAPTER 20
d.
XL = 2 fL  2 (1.027 MHz)(200  H) = 1.291 k
86.04 0)(114.1 V)
VR =
= 7.587 V
86.04   j1.291 k
P = VR2 / R  (7.587 V)2 86.04  = 669 mW
13.69 W: 669 mW  20:1
16.
a.
Q =

X L 100 
= 5  10
=
RL
20 
1
R2 + X 2 (20 )2 + (100 ) 2
XL

X
=
= 104 
=
=
C
100 
XL
R2 + X L2 X C
2
+ X 2L
10,400 
= 342.11 
= 1000  
20
R
b.
ZT = Rs  Rp = Rs  R 
c.
E = IZTp = (5 mA 0)(342.11  0) = 1.711 V 0
IC =
E
1.711 V 0
= 16.45 mA 90
=
X C   90 104    90
ZL = 20  + j100  = 101.98  78.69
E
1.711 V 0
= 16.78 mA 78.69
IL =
=
Z L 101.98  78.69
d.
e.
17.
100 
= 795.77 H
L= XL =
2 f 2 (20 kHz)
1
1
= 76.52 nF
C=
=
2 fX C 2 (20 kHz)(104 )
R 342.11 
= 3.29
=
XC
104 
BW = fp/Qp = 20,000 Hz/3.29 = 6079.03 Hz
Qp =
 2  106

2 
Hz  (1 mH)
 2

X
X
2000 

= 57.14 
Q  35  L  R  L 
35
35
35
R
Q  10 : Q p 
CHAPTER 20
fp
BW

2  106 / 2 Hz
 20
100,000 Hz
271
R  Q2 R R  (35) 2 57.14 

Q p  20 
2000
XL
And 40,000 = R  70, 000
So R = 93.33 k  use R = 91 k (standard value)
1
1
Q p  10, X C  X L  2000  

2 fC
 2  106

2 
Hz  C
 2

C = 250 pF  use C = 240 pF (standard value)
18.
a.
fs =
1
2 LC
=
1
= 102.73 kHz
2 (80  H)(0.03  F)
2
(1.5 )2 0.03  F
C
fp = fs 1  R  = 102.73 kHz 1 
= 102.73 kHz(.99958)
L
80  H
= 102.69 kHz
1  2C 
fm = fs 1   R   = 102.73 kHz(0.99989) = 102.72 kHz
4 L 
Since fs  fp  fm  high Qp
b.
XL = 2πfpL = 2π(102.69 kHz)(80 μH) = 51.62 
1
1
=
= 51.66 
XC =
2 f p C 2 (102.69 kHz)(0.03  F)
XL  XC
c.
ZTp = Rs  Q2 R
51.62 
Q = X L =
= 34.41
R
1.5 
ZTp = 10 k   (34.41) 21.5  = 10 k   1.776 k  = 1.51 k
2
R s  Q  R  Z T p 1.51 k 
= 29.25
=
=
XL
X L 51.62 
f
102.69 kHz
BW = p =
= 3.51 kHz
29.25
Qp
d.
Qp =
e.
Converting the voltage source to a current source:
E 100 V
= 10 mA
Is =

Rs 10 k
And Rs = Rp = 10 k
R s I s = 10 k (10 mA) = 8.49 mA
Then IT =
2
R s + Q  R  10 k  + 1.78 k 
IC = IL  Q IT = (34.41)(8.49 mA) = 292.14 mA
272
CHAPTER 20
19.
f.
VC = IZTp = (10 mA)(1.51 k) = 15.1 V
a.
fs =
1
2 LC
=
1
= 7.12 kHz
2 (0.5 mH)(1  F)
R2 C
(8 Ω) 2 (1  F)
= 7.12 kHz 1 
= 7.12 kHz(0.9338) = 6.65 kHz
0.5 mH
L
fp = fs 1 
1  R 2C 
1  (8 Ω) 2 (1  F) 
fm = fs 1     = 7.12 kHz 1  
 = 7.12 kHz (0.9839)
4  0.5 mH 
4 L 
= 7.01 kHz
Low Qp
b.
XL = 2πfpL = 2π(6.647 kHz)(0.5 mH) = 20.88 
1
1
=
= 23.94 
XC =
2 fC 2 (6.647 kHz)(1  F)
XC > XL (low Q)
c.
ZTp = Rs  Rp = Rs 
R  + X L = 500   (8 ) + (20.88 ) = 500   62.5 
8
R
2
2
2
2
= 55.56 
d.
ZT p
Qp =
55.56 
= 2.32
23.94 
=
6.647 kHz
= 2.87 kHz
2.32
X Lp
BW =
e.
=
fp
Qp
One method: VC = IZTp = (40 mA)(55.56 ) = 2.22 V
IC =
IL =
20.
VC
2.22 V
= 92.73 mA
=
X C 23.94 
VC
2.22 V
2.22 V
=
= 99.28 mA
=
R  + jX L 8 + j 20.88 22.36 
f.
VC = 2.22 V
a.
ZT p =
R2 + X L2
= 50 k
R
(50 )2 + X L2 = (50 k)(50 )
XL =
CHAPTER 20
250  104  2.5  103 = 1580.3 
273
b.
Q=
X L 1580.3
= 31.61  10
=
50
R
 XC = XL = 1580.3 
21.
XL
1580.3 Ω
= 15.72 kHz
=
2 L 2π(16 mH)
c.
XL = 2πfpL  fp =
d.
XC =
a.
Q = 20 > 10  fp = fs =
b.
Q =
1
1
1
 C=
= 6.4 nF
=
2 f p C
2 f s X C 2 (15.72 kHz)(1580.3 )
1
2 LC
=
1
= 3558.81 Hz
2 (200 mH)(10 nF)
X L 2 fL
2 fL 2 (3558.81 Hz)(0.2 H)
 R =
= 223.61 
=
=
R
R
20
Q
ZTp = Rs  R p = Rs  Q2 R = 40 k  (20)2 223.61 
ZTp = 27.64 k
Converting the voltage source to a current source:
E 200 V
= 5 mA
Is 

Rs 40 k
Rp = Rs = 40 k
VC = IZTp = (5 mA)(27.64 k) = 138.2 V
c.
P = I2R = (5 mA)227.64 k = 691 mW
d.
Qp =
BW =
22.
fp
Qp
=
3558.81 Hz
= 575.86 Hz
6.18
a.
Ratio of XC to R suggests high Q system.
 XL = 400  = XC
b.
Q =
c.
Qp =
BW =
274
27.64 k 
R Rs  R p
= 6.18
=
=
2 (3558.81 Hz)(0.2 H)
XL
XL
X L 400 
= 50
=
R
8
R
XL
fp
Qp
=
Rs  R p
XL
=
Rs  Q 2R
XL
=
20 k   (50)2 8  10 k 
=
= 25
400 
400 
 fp = QpBW = (25)(1000 Hz) = 25 kHz
CHAPTER 20
23.
d.
V C max = IZTp = (0.1 mA)(10 k) = 1 V
e.
f2 = fp + BW/2 = 25 kHz +
a.
XC =
1 kHz
= 25.5 kHz
2
1 kHz
f1 = fp  BW/2 = 25 kHz 
= 24.5 kHz
2
R2 + X L2
 X L2  XLXC + R2 = 0
XL
X L2  100 XL + 144 = 0
XL =
(  100)  (100) 2  4(1)(144)
2
104  576
= 50   48.54 
2
XL = 98.54  or 1.46 
= 50  
b.
c.
Q =
Qp =
X L 98.54 
= 8.21
=
R
12 
Rs  R p
X Lp
R2 + X L2
(12 ) 2 + (98.54 ) 2
40 k 
R
12 
=
XC
100 
40 k 
=
40 k  821.18  804.66 
=
= 8.05
100 
100 
BW = fp/Qp  fp = QpBW = (8.05)(1 kHz) = 8.05 kHz
=
24.
d.
VCmax = IZTp = (6 mA)(804.66 ) = 4.83 V
e.
f2 = fp + BW/2 = 8.05 kHz +
a.
1 kHz
= 8.55 kHz
2
1 kHz
f1 = fp  BW/2 = 8.05 kHz 
= 7.55 kHz
2
1
1
fs =
=
= 41.09 kHz
2 LC 2 (0.5 mH)(30 nF)
fp = fs 1 
R2 C
(6 )2 30 nF
= 41.09 kHz 1 
= 41.09 kHz(0.9978) = 41 kHz
0.5 mH
L
1  R 2C 
1  (6 ) 2(30 nF) 
fm = fs 1     = 41.09 kHz 1  
 = 41.09 kHz(0.0995)
4  0.5 mH 
4 L 
= 41.07 kHz
High Qp
CHAPTER 20
275
b.
80 V 0
= 4 mA 0, Rs = 20 k
20 k  0
X
2 fL 2 (41 kHz)(0.5 mH)
Q = L =
= 21.47 (high Q coil)
=
R
R
6
I=
R + X L
(6 ) 2 + (128.81 ) 2
20
k


Rs  R p
R =
6
=
Qp =
2
2
(6 ) 2 + (128.81 ) 2
R + X L
X Lp
128.81 
XL
20 k   2.771 k  2.434 k 
=
= 18.86 (high Qp)
=
129.09 
129.09 
Rs 
2
c.
ZTp = Rs  Rp = 20 k  2.771 k = 2.43 k
d.
VC = IZTp = (4 mA)(2.43 k) = 9.74 V
e.
BW =
f.
XC =
fp
Qp
Q = X L =
R
41 kHz
= 2.17 kHz
18.86
=
1
1
=
= 129.39 
2 fC 2 (41 kHz)(30 nF)
V
9.736 V
= 75.25 mA
IC = C =
X C 129.39 
IL =
25.
2
9.736 V
9.736 V
VC
=
= 75.50 mA
=
R + jX L 6  + j128.81  128.95 
2 f p L
R
 R =
2 f p L
Q

2 (20 kHz)(2 mH)
= 3.14 
80
BW = fp/Qp  Qp = fp/BW = 20 kHz/1.8 kHz = 11.11
1
1
1
C=
= 2
= 31.66 nF
High Q fp  fs =
2 2
4 f p L 4 (20 kHz) 2 2 mH
2 LC
Qp =
R
XC
 R = QpXC =
Qp
2 f p C
=
11.11
= 2.79 k
2 (20 kHz)(31.66 nF)
= (80) 3.14  = 20.1 k
Rp R
(20.1 k)(2.793 k)
RsR p
R = Rs  Rp =
 Rs =
= 3.24 k
=
R p  R 20.1 k  2.793 k
Rs + R p
Rp = Q2 R
276
2
CHAPTER 20
26.
VCmax
1.8 V
= 9 k
I
0.2 mA
R p 9 k
R Rs  R p R p
Qp =
 XL =
=
= 300  = XC
=
=
XL
XL
XL
30
Qp
VCmax  IZTp  ZTp =
BW =
fp
Qp
 fp = QpBW = (30)(500 Hz) = 15 kHz
L=
XL
300 
= 3.18 mH
=
2 f 2 (15 kHz)
C=
1
1
= 35.37 nF
=
2 fX C 2 (15 kHz)(300 )
Qp = Q (Rs=  ) =
27.
a.
=
fs =
1
2 LC
Q =
=
X
300 Ω
XL
 R = L =
= 10 
Qp
30
R
1
2 (200  H)(2 nF)
= 251.65 kHz
X L 2 (251.65 kHz)(200  H)
= 15.81  10
=
R
20 
 fp = fs = 251.65 kHz
b.
ZTp = Rs  Q2 R = 40 kΩ  (15.81)2 20  = 4.44 k
c.
Qp =
d.
BW =
e.
20 μH, 20 nF
fs the same since product LC the same
fs = 251.65 kHz
X
2 (251.65 kHz)(20  H)
Q = L =
= 1.581
20 
R
Low Q :
Rs  Q 2 R  4.444 k
=
= 14.05
XL
316.23 
fp
Qp
=
251.65 kHz
= 17.91 kHz
14.05
fp = fs 1 
R2 C
(20 ) 2 (20 nF)
= (251.65 kHz) 1 
20  H
L
= (251.65 kHz)(0.775) = 194.93 kHz
XL = 2πfpL = 2π(194.93 kHz)(20 μH) = 24.496 Ω
R 2 + X L2 (20 )2 + (24.496 ) 2
= 50 
Rp = 
=
20 
R
ZTp = Rs  Rp = 40 k  50  = 49.94 
CHAPTER 20
277
R
49.94 
= 2.04
=
X L 24.496 
f
194.93 kHz
BW = p =
= 95.55 kHz
2.04
Qp
Qp =
f.
0.4 mH, 1 nF
fs = 251.65 kHz since LC product the same
2 (251.65 kHz)(0.4 mH)
Q = X L =
= 31.62  10
20 
R
 fp = fs = 251.65 kHz
ZTp = Rs  Q2 R = 40 k  (31.62)2 20  = 40 k  ( 20 k)  13.33 k
Rs  Q2 R 13.33 k 
= 21.08
=
632.47 
XL
f p 251.65 kHz
=
= 11.94 kHz
BW =
Qp
21.08
Qp =
g.
h.
L 200  H
= 100  103
=
C
2 nF
L 20  H
part (e) =
= 1  103
C 20 nF
L 0.4 mH
part (f) =
= 400  103
C
1 nF
Network
L
ratio increased BW decreased.
C
Also, Vp = IZTp and for a fixed I, ZTp and therefore Vp will increase with increase in the
Yes, as
L/C ratio.
278
CHAPTER 20
Chapter 21
1.
3
 = 0.1875, d2 = 1
16
Value = 103  100.1875"/1"
= 103  1.54
= 1.54 kHz
3
right: d1 =  = 0.75, d2 = 1
4
Value = 103  100.75"/1"
= 103  5.623
= 5.62 kHz
left: d1 =
a.
b.
bottom:
top:
5
15
 = 0.3125, d2 =
 = 0.9375
16
16
Value = 101  100.3125"/ 0.9375" = 101  100.333
= 101  2.153
= 0.22 V
11
 = 0.6875, d2 = 0.9375
d1 =
16
Value = 101  100.6875"/ 0.9375" = 101  100.720
= 101  5.248
= 0.52 V
d1 =
a.
5
b.
4
c.
8
d.
6
e.
1.30
f.
3.94
g.
4.75
h.
0.498
a.
1000
b.
1012
c.
1.59
d.
1.1
e.
1010
f.
1513.56
g.
10.02
h.
1,258,925.41
4.
a.
11.51
b.
−9.21
c.
2.996
5.
log10 48 = 1.68
log10 8 + log10 6 = 0.903 + 0.778 = 1.68
6.
log10 0.2 = 0.699
log10 18  log10 90 = 1.255  1.954 = 0.699
7.
log10 0.5 = 0.30
log10 2 = (0.301) = 0.30
8.
log10 27 = 1.43
3 log10 3 = 3(0.4771) = 1.43
2.
3.
CHAPTER 21
d.
9.07
279
9.
280 mW
P2
= log10
= log10 70 = 1.85
4 mW
P1
a.
bels = log10
b.
dB = 10 log10
P2
= 10(log10 70) = 10(1.845) = 18.45
P1
10.
dB = 10 log10 P 2
P1
100 W
6 dB = 10 log10
P1
0.6 = log10 x
100 W
x = 3.981 =
P1
100 W
P1 =
= 25.12 W
3.981
11.
dB = 10 log10
12.
dBm = 10 log10
13.
dBv = 20 log10
8.4 V
V2
= 20 log10
= 20 log 10 84 = 38.49
0.1 V
V1
14.
dBυ = 20 log10
V2
V1
22 = 20 log10
Vo
20 mV
40 W
P2
= 10 log10
= 10 log10 20 = 13.01
2W
P1
P
1 mW
120 mW
= 10 log10 120 = 20.79
dBm = 10 log10
1 mW
1.1 = log10 x
Vo
20 mV
Vo = 251.79 mV
x = 12.589 =
15.
280
P
0.0002  bar
0.001  bar
dBs = 20 log10
= 13.98
0.0002  bar
0.016  bar
dBs = 20 log10
= 38.06
0.0002  bar
Increase = 24.08 dBs
dBs = 20 log10
CHAPTER 21
16.
60 dBs  90 dBs
quiet
loud
60 dBs = 20 log10
P1
= 20 log10x
0.002  bar
3 = log10x
x = 1000
90 dBs = 20 log10
P2
= 20 log10y
0.002  bar
4.5 = log10y
y = 31.623  103
P1
x 0.002  bar
103
=
= P1 =
y
P2
P 2 31.623  103
0.002  bar
and P2 = 31.62 P1
18.
b.
19.
V2
0.775 V
V2
0.4 = log10
0.775 V
a.
a.
8 dB = 20 log10
V2
= 2.512
0.775 V
V2 = (2.512)(0.775 V) = 1.947 V
2
(1.947 V)2
= 6.32 mW
P= V =
R
600 
V2
5 dB = 20 log10
0.775 V
V2
0.25 = log10
0.775 V
V2
= 0.562
0.775 V
V2 = (0.562)(0.775 V) = 0.436 V
2
(0.436 V) 2
V
=
= 0.32 mW
P=
600 
R
Aυ =
Vo
=
Vi
XC
R
2
 X C2
90 + tan1 XC/R =
1
2
 R 

 1
 XC 
tan1 R/XC
1
1
= 3617.16 Hz
=
2 RC 2 (2.2 k)(0.02  F)
V
f = f c:
Aυ = o = 0.707
Vi
fc =
CHAPTER 21
281
f = 0.1fc:
At fc, XC = R = 2.2 k
1
1
1  1 
XC =
=
=

 = 10  2.2 k   = 22 k
2 fC 2 0.1 f c C 0.1  2 f c C 
Aυ =
f = 0.5fc =
1
fc :
2
XC =
Aυ =
f = 2fc:
XC =
Aυ =
f = 10fc:
XC =
Aυ =
b.
θ = tan1 R/XC
f = f c:
f = 0.1fc:
f = 0.5fc:
f = 2fc:
f = 10fc:
20.
a.
1
=
2
 R 

 1
 XC 
1
2
 2.2 k  

 +1
 22 k  
=
1
(.1) 2 + 1
= 0.995
 1 
1
= 2
 = 2  2.2 k  = 4.4 k
2 f cC 
 fc 

2   C
 2
1
1
=
= 0.894
2
(0.5) 2 + 1
 2.2 k  

 +1
 4.4 k  
1
=
2 fC
1
1 1  1
= 
 =  2.2 k   = 1.1 k
2 (2f c )C 2  2 f cC  2
1
2
 2.2 k  

 +1
 1.1 k  
=
1
(2) 2 + 1
= 0.447
1
1  1  1
= 
 =  2.2 k   = 0.22 k
2 (10 f c )C 10  2 f cC  10
1
2
 2.2 k  

 +1
 0.22 k  
=
1
(10) 2 + 1
= 0.0995
θ = tan1 = 45
1
= 5.71
10
1
θ = tan1 2.2 k/4.4 k = tan1 = 26.57
2
θ = tan1 2.2 k/1.1 k = tan1 2 = 63.43
θ = tan1 2.2 k/0.22 k = tan1 10 = 84.29
θ = tan1 2.2 k/22 k = tan1
1
1
=
= 15.915 kHz
2 RC 2 ( 1 k )(0.01  F)
f = 2fc = 31.83 kHz
1
1
XC =
=
= 500 
2 fC 2 (31.83 kHz)(0.01  F)
V
500 
XC
Aυ = o =
= 0.4472
=
2
2
Vi
(1 k )2 + (0.5 k ) 2
R X
fc =
C
Vo = 0.4472Vi = 0.4472(10 mV) = 4.47 mV
282
CHAPTER 21
b.
1
1
f c = (15,915 kHz) = 1.5915 kHz
10
10
1
1
XC =
=
= 10 k
2 fC 2 (1.5915 kHz)(0.01  F)
V
10 k 
XC
=
= 0.995
Aυ = o =
2
2
Vi
(1 k ) 2 + (10 k ) 2
R + XC
f=
Vo = 0.995Vi = 0.995(10 mV) = 9.95 mV
c.
21.
Yes, at f = fc, Vo = 7.07 mV
1
f c , Vo = 9.95 mV (much higher)
at f =
10
at f = 2fc, Vo = 4.47 mV (much lower)
fc = 500 Hz =
1
1
=
2 RC 2 (1.2 k )C
1
1
= 0.265 μF
=
2 Rfc 2 (1.2 k )(500 Hz)
V
1
Aυ = o =
2
Vi
 R 

 1
 XC 
C=
At f = 250 Hz, XC = 2402.33  and Aυ = 0.895
At f = 1000 Hz, XC = 600.58  and Aυ = 0.4475
θ = tan1R/XC
1
At f = 250 Hz = fc, θ = 26.54
2
At f = 1 kHz = 2fc, θ = 63.41
CHAPTER 21
283
22.
1
1
=
= 67.73 kHz
2 RC 2 (4.7 k )(500 pF)
a.
fc =
b.
f = 0.1 fc = 0.1(67.726 kHz)  6.773 kHz
1
1
=
= 46.997 k
XC =
2 fC 2 (6.773 kHz)(500 pF)
V
46.997 k 
XC
=
Aυ = o =
= 0.995  1
2
2
Vi
R +X
(4.7 k )2 + (46.997 k ) 2
C
c.
f = 10fc = 677.26 kHz
1
1
=
 470 
XC =
2 fC 2 (677.26 kHz)(500 pF)
V
470 
XC
=
Aυ = o =
= 0.0995  0.1
2
2
Vi
R + XC
(4.7 k ) 2 + (470 ) 2
d.
Aυ =
Vo
= 0.01 =
Vi
XC
2
R + X C2
X C = 100 X
C
0.01
R2 + X C2 = 104 X C2
2
2
R +XC=
R2 = 104 X C2  X C2 = 9,999 X C2
R
4.7 k 
 47 
XC =
=
9,999 99.995
XC =
23.
a.
Aυ =
1
1
1
f=
=
= 6.77 MHz
2 fC
2 X CC 2 (47 )(500 pF)
R
Vo
=
tan1 XC/R =
2
2
Vi
R  XC
fc =
f = f c:
f = 2fc:
X 
1+  C 
 R 
2
tan1 XC/R
1
1
= 3.62 kHz
=
2 RC 2 (2.2 k)(0.02  F)
Vo
= 0.707
Vi
At fc, XC = R = 2.2 k
1
1
1 1  1
XC =
=
= 
 =  2.2 k  = 1.1 k
2 fC 2 (2 f c )C 2  2 f cC  2
Aυ =
Aυ =
284
1
1
 1.1 k  
1+ 

 2.2 k  
2
= 0.894
CHAPTER 21
f=
1
f c:
2
f = 10fc:
 1 
1
= 2
 = 2  2.2 k  = 4.4 k
f
 
 2 f cC 
2  c  C
 2
1
Aυ =
= 0.447
2
 4.4 k  
1+ 

 2.2 k  
1
1  1  2.2 k 
XC =
= 
= 0.22 k
=
2 (10 f c )C 10  2 f cC 
10
XC =
Aυ =
f=
b.
1
f c:
10
= 0.995
2
 0.22 k  
1+ 

 2.2 k  
 1 
1
XC =
= 10 
 = 10  2.2 k   = 22 k
 f 
 2 f cC 
2  c  C
 10 
1
Aυ =
= 0.0995
2
22
k



1+ 

2.2
k



f = f c,
θ = 45
f = 2fc,
θ = tan1 (XC/R) = tan1 1.1 k/2.2 k = tan1
f=
1
f c,
2
f = 10fc,
f=
24.
1
1
f c,
10
a.
f = fc: Aυ =
b.
fc =
1
= 26.57
2
4.4 k 
= tan1 2 = 63.43
2.2 k 
0.22 k 
θ = tan1
= 5.71
2.2 k 
22 k 
θ = tan1
= 84.29
2.2 k 
θ = tan1
Vo
= 0.707
Vi
1
1
=
= 15.915 kHz
2 RC 2 (10 k )(1000 pF)
f = 4fc = 4(15.915 kHz) = 63.66 kHz
1
1
=
= 2.5 k
XC =
2 fC 2 (63.66 kHz)(1000 pF)
V
R
10 k 
=
Aυ = o =
= 0.970 (significant rise)
2
2
Vi
R  XC
(10 k )2 + (2.5 k ) 2
CHAPTER 21
285
25.
c.
f = 100fc = 100(15.915 kHz) = 1591.5 kHz  1.592 MHz
1
1
=
= 99.972 
XC =
2 fC 2 (1.592 MHz)(1000 pF)
R
10 k 
=
Aυ =
= 0.99995  1
2
2
(10 k ) 2 + (99.972 ) 2
R + XC
d.
At f = fc, Vo= 0.707Vi = 0.707(10 mV) = 7.07 mV
2
(7.07 mV ) 2
Po = V o =
 5 nW
R
10 k 
Aυ =
Vo
=
Vi
fc =
1
1
1
 R=
= 795.77 
=
2 RC
2 f cC 2 ( 2 kHz)(0.1  F)
1
X 
1+  C 
 R 
2
tan1 XC/R
R = 795.77   750

 47 
 = 797 

nominal values
1
= 1996.93 Hz using nominal values
 fc =
2 (797 )(0.1  F)
At
286
f = 1 kHz, Aυ = 0.458
f = 4 kHz, Aυ  0.9
θ = tan1 X C
R
f = 1 kHz, θ = 63.4
f = 4 kHz, θ = 26.53
CHAPTER 21
26.
1
1
=
= 79.58 kHz
2 RC 2 (100 k )(20 pF)
a.
fc =
b.
f = 0.01fc = 0.01(79.577 kHz) = 0.7958 kHz  796 Hz
1
1
=
= 9.997 M
XC =
2 fC 2 (796 Hz)(20 pF)
R
100 k 
Aυ = V o =
=
= 0.01  0
2
2
Vi
(100 k ) 2 + (9.997 M ) 2
R + XC
c.
f = 100fc = 100(79.577 kHz)  7.96 MHz
1
1
=
= 999.72 
XC =
2 fC 2 (7.96 MHz)(20 pF)
V
R
100 k 
=
Aυ = o =
= 0.99995  1
2
2
Vi
(100 k ) 2 + (999.72 ) 2
R + XC
d.
Aυ =
Vo
R
= 0.5 =
Vi
R 2 + X C2
R 2 + X C2 = 2R
R2 + X C2 = 4R2
X C2 = 4R2  R2 = 3R2
3R 2 = 3R = 3 (100 k) = 173.2 k
1
1
1
f=
XC =
=
2 fC
2 X C C 2 (173.2 k )(20 pF)
f = 45.95 kHz
XC =
27.
a.
1
1
=
= 795.77 Hz
2 RC 2 (0.1 k )(2  F)
1
1
= 1.94 Hz
f c2 =
=
2 RC 2 (10 k )(8200 pF)
low-pass section:
f c1 =
high-pass section:
For the analysis to follow, it is assumed (R2 + jX C 2 )  R1  R1 for all frequencies of
interest.
At f c1 = 795.77 Hz:
VR1 = 0.707 Vi
X C2 =
|Vo | =
1
= 24.39 k
2 fC2
24.39 k (VR1 )
(10 k ) 2 + (24.39 k ) 2
= 0.925 VRi
Vo = (0.925)(0.707 Vi) = 0.654 Vi
CHAPTER 21
287
At f c2 = 1.94 kHz:
Vo = 0.707 VR1
1
= 41 
2 fC1
R1Vi
100 (Vi )
= 0.925 Vi
VR1 =
=
(100 ) 2 + (41 ) 2
R12 + X C21
X C1 =
|Vo | = (0.707)(0.925 Vi) = 0.64 Vi
(1.94 kHz  795.77 Hz)
= 1.37 kHz
2
X C 1 = 58.1 , X C 2 = 14.17 k
At f = 795.77 Hz +
VR1 =
Vo =
100 (Vi )
(100 ) 2 + (58.1 )2
= 0.864 Vi
 
14.17 k  VR1
(10 k ) 2 + (14.17 k ) 2
= 0.817 VR1
Vo = 0.817(0.864 Vi) = 0.706Vi
V
and Aυ = o = 0.706 ( maximum value)
Vi
After plotting the points it was determined that the gain should also be determined at
f = 500 Hz and 4 kHz:
f = 500 Hz:
X C1 = 159.15 , X C2 = 38.82 k,
VR1 = 0.532 Vi, Vo = 0.968 VR1
f = 4 kHz:
Vo = 0.515 Vi
X C1 = 19.89 , X C2 = 4.85 kΩ,
VR1 = 0.981 Vi, Vo = 0.437 VR1
Vo = 0.429 Vi
b.
288
Using 0.707(.706)  0.5 to define the bandwidth
BW  3.4 kHz  0.48 kHz = 2.92 kHz
and BW  2.9 kHz
 2.9 kHz 
with fcenter = 480 Hz + 
 = 1930 Hz
2 

CHAPTER 21
28.
1
= 4 kHz
2 R1C1
Choose R1 = 1 k
1
1
= 39.8 nF  Use 39 nF
C1 =
=
2 f1R1 2 (4 kHz)(1 k )
f1 =
1
= 80 kHz
2 R2C2
Choose R2 = 20 k
1
1
= 99.47 pF  Use 100 pF
C2 =
=
2 f 2 R2 2 (80 kHz)(20 k )
f2 =
80 kHz  4 kHz
= 42 kHz
2
At f = 42 kHz, X C1 = 97.16 , X C2 = 37.89 k
Center frequency = 4 kHz +
Assuming Z2  Z1
R1 (Vi )
= 0.995Vi
|VR1 |=
R12 + X C21
|Vo | =
X C2 (VR1 )
R22 + X C21
= 0.884Vi
Vo = 0.884 VR1 = 0.884(0.995Vi) = 0.88 Vi
as f = f1: VR1 = 0.707Vi, X C2 = 221.05 k
and Vo = 0.996 VR1
so that Vo = 0.996 VR1 = 0.996(0.707Vi) = 0.704Vi
Although Aυ = 0.88 is less than the desired level of 1, f1 and f2 do define a band of frequencies
for which Aυ  0.7 and the power to the load is significant.
CHAPTER 21
289
29.
a.
b.
c.
fs =
1
2 LC
=
1
= 98.1 kHz
2 (4.7 mH)(560 pF)
XL
2 (98.1 kHz)(4.7 mH)
= 16.84
=
160  + 12 
R + R
f
98.1 kHz
BW = s =
= 5.83 kHz
16.84
Qs
Qs =
R
160 (1 V)
V
= 0.93 V and Aυ = o = 0.93
Vi=
Vi
R  R
172 
5.83 kHz
BW
f1 = fs 
= 98.1 kHz 
= 95.19 kHz
Since Qs  10,
2
2
BW
f2 = fs +
= 101.02 kHz
2
At f = 95.19 kHz: XL = 2πfL = 2π(95.19 kHz)(4.7 mH) = 2.81 k
1
1
= 2.99 k
XC =
=
2 fC 2 (95.19 kHz)(560 pF)
160 (1 V 0)
160 V 0
Vo =
=
172 + j 2.81 k   j 2.99 k  172  j180
160 V 0
= 0.643 V46.30
=
248.97   46.30
At f = fs: Vomax =
At f = 101.02 kHz: XL = 2πfL = 2π(101.02 kHz)(4.7 mH) = 2.98 k
1
1
= 2.81 k
XC =
=
2 fC 2 (101.02 kHz)(560 pF)
160 (1 V 0)
160 V 0
Vo =
=
172 + j 2.98 k   j 2.81 k  172 + j170
160 V 0
= 0.66 V44.66
=
241.83 44.66
d.
f = fs: Vomax = 0.93 V
f = f1 = 95.19 kHz, Vo = 0.707(0.93 V) = 0.66 V
f = f2 = 101.02 kHz, Vo = 0.707(0.93 V) = 0.66 V
30.
a.
R2C
 159.15 kHz
L
2 LC
2 f p L 2 (159.15 kHz)(1 mH)
X
Q = L =
= 62.5  10
=
16 
R
R
fp =
1
1
ZTp = Q2 R = (62.5)2 16  = 62.5 k  4 k
and Vo  Vi at resonance.
290
CHAPTER 21
However, R = 3.3 k affects the shape of the resonance curve and BW = fp/ Q cannot be
applied.
V
For Aυ = o = 0.707, | X | = R for the following configuration
Vi
For frequencies near fp, XL  R and ZL = R + jXL  XL
and X = XL  XC.
For frequencies near fp but less than fp
XC X L
X=
XC  X L
and for Aυ = 0.707
XC X L
=R
XC  X L
1
and XL = 2πf1L
2 f1C
the following equation can be derived:
Substituting XC =
f12 +
1
1
f1  2
=0
2 RC
4 LC
For this situation:
1
1
= 48.23  103
=
2 RC 2 (3.3 k )(0.001  F)
1
1
= 2.53  1010
= 2
2
4 LC 4 (1 mH)(0.001  F)
and solving the quadratic equation, f1 = 135.83 kHz
BW
= fp  f1 = 159.15 kHz  135.83 kHz = 22.32 kHz
and
2
BW
so that f2 = fp +
= 159.15 kHz + 18.75 kHz = 177.9 kHz
2
b.
fp
159.15 kHz
= 3.57
BW 44.64 kHz
BW = 2(18.75 kHz) = 37.5 kHz
Qp =
CHAPTER 21
=
291
31.
b.
a.
5000 
5000 
XL
= 12.5
=
=
R + R 390  + 10  400 
Qs =
f s 5000 Hz
= 400 Hz
=
12.5
Qs
400 Hz
= 4.8 kHz
f1 = 5000 Hz 
2
400 Hz
= 5.20 kHz
f2 = 5000 Hz +
2
BW =
c.
At resonance
10 (Vi )
Vo =
10  + 400 
= 0.024 Vi
32.
d.
At resonance,
a.
Q =
10 Ω  2 k = 9.95 Ω
9.95 (Vi )
 0.024 Vi as above!
Vo =
9.95  + 400 
X L 400 
= 40
=
R
10 
Z T p = Q  R = (40) 20  = 32 k  1 k
2
2
At resonance, Vo =
32 k  Vi
= 0.97Vi
32 k  + 1 k 
Vo
= 0.97
Vi
For the low cutoff frequency note solution to Problem 30:
1
1
f12 +
f1  2
=0
2 RC
4 LC
1
1
C=
= 19.9 nF
=
2 fX C 2 (20 kHz)(400 )
and Aυ =
L=
XL
400 
=
= 3.18 mH
2 f 2 (20 kHz)
Substituting into the above equation and solving
f1 = 16.4 kHz
BW
= 20 kHz  16.4 kHz = 3.6 kHz
with
2
and BW = 2(3.6 kHz) = 7.2 kHz
f
20 kHz
= 2.78
Qp = p =
BW 7.2 kHz
292
CHAPTER 21
b.

c.
At resonance
Z T p = 32 k  100 k = 24.24 k
24.24 k  V i
= 0.96Vi
24.24 k  + 1 k 
V
and Aυ = o = 0.96 vs 0.97 above
Vi
with Vo =
At frequencies to the right and left of fp, the impedance Z T p will decrease and be
affected less and less by the parallel 100 k load. The characteristics, therefore, are
only slightly affected by the 100 k load.
d.
At resonance
Z T p = 32 k  20 k = 12.31 k
with Vo =
12.31 k  Vi
= 0.925Vi vs 0.97 Vi above
12.31 k  + 1 k 
At frequencies to the right and left of fp, the impedance of each frequency will actually
be less due to the parallel 20 k load. The effect will be to narrow the resonance curve
and decrease the bandwidth with an increase in Qp.
33.
a.
fp =
1
2 LC
1
= 726.44 kHz (band-stop)
2 (400  H)(120 pF)
=


X Ls 90 + X L p 90  X C   90 = 0
jX Ls +
jX Ls +
 X 90  X   90 = 0
C
Lp
jX L p  jX C

X Lp X C

=0
X  X 
=0
j X Lp  X C
jX Ls  j
X Lp X C
Lp
X Ls 
X Lp X C
X Lp  X C
C
=0
X Ls X C  X Ls X L p + X L p X C = 0
 Lp
 Ls
  Ls +
=0
C
C
CHAPTER 21
293
1
 Ls + L p  = 0
C
Ls + L p
=
CLs L p
LsLpω2 
f=
34.
a.
Ls + L p
1
2
CLs L p
1
=
460  106
= 2.01 MHz (pass-band)
28.8  1019
1
2
1
1
= 12.68 mH
= 2
2
4 f s C 4 (100 kHz ) 2(200 pF)
2 LC
XL = 2πfL = 2π(30 kHz)(12.68 mH) = 2388.91 
1
1
=
= 26.54 k
XC =
2 fC 2 (30 kHz)(200 pF)
XC  XL = 26.54 k  2388.91  = 24.15 k(C)
X Lp = X C(net) = 24.15 k
fs =
Lp =
 Ls =
2
XL
24.15 k 
= 128.19 mH
=
2 f 2 (30 kHz)
35. a, b.
fc =
1
1
= 7.2 kHz
=
2 RC 2 (0.47 k )(0.047  F)
c.
f=
1
f c:
2
A dB = 20 log10
f = 2fc:
A dB = 20 log10
1
f c:
10
A dB = 20 log10
f = 10fc:
A dB = 20 log10
f=
d.
294
1
f c:
2
Aυ =
f = 2fc:
Aυ =
f=
1
1+  f c /f 
1
1
1  (10)2
1 + (0.1) 2
1
1  (0.5) 2
=
1
1+ (2)2
= 7 dB
= 20.04 dB
1
2
= 20 log10
= 0.969 dB
1  (0.5) 2
1
1 +  f c /f 
2
= 0.043 dB
1
1 + (2) 2
= 0.447
= 0.894
CHAPTER 21
e.
36.
a.
fc =
1
1
1
= 1.83 kHz
=
=
2 RC 2 (6.8 k   12 k )0.02  F 2 (4.34 k )(0.02  F)
Vo
1
=
Vi
1 + ( f /f ) 2
c


1
Vi
and Vo = 
 1 + ( f /f ) 2 
c


b.
c. & d.
CHAPTER 21
295
e.
Remember the log scale! 1.5fc is not midway between fc and 2fc
A dB = 20 log10 Aυ
1.5 = 20 log10 Aυ
0.075 = log10 Aυ
V
Aυ = o = 0.84
Vi
f.
37. a, b.
θ = tan1 fc/f
Aυ =
fc =
c.
Vo
= A  =
Vi
1
1 + (f/f c )
2
tan1f/fc
1
1
=
= 13.26 kHz
2 RC 2 (12 k )(1 nF)
f = fc/2 = 6.63 kHz
AdB  20 log10
1
1 + (0.5) 2
= 0.97 dB
f = 2fc = 26.52 kHz
AdB = 20 log10
1
1 + (2) 2
= 6.99 dB
f = fc/10 = 1.326 kHz
AdB = 20 log10
1
1 + (0.1) 2
= 0.04 dB
f = 10fc = 132.6 kHz
A dB = 20 log10
d.
296
f = fc/2:
Aυ =
f = 2fc:
Aυ =
1
1 + (10) 2
1
1  (0.5) 2
1
1 + (2) 2
= 20.04 dB
= 0.894
= 0.447
CHAPTER 21
38.
e.
θ = tan1 f/fc
θ = tan1 0.5 = 26.57
f = fc/2:
θ = tan1 1 = 45
f = f c:
θ = tan1 2 = 63.43
f = 2fc:
a.
R2  XC =
( R2 )(  jX C )
R2 X C
= j
R2  jX C
R2  jX C
  jR2 X C 

 Vi
R2  jX C 
R2 X C Vi

Vo =
= j
R2 X C
R1 ( R2  jX C )  jR2 X C
R1  j
R2  jX C
 jR2 X C Vi
-jR2 X C Vi
=
=
R1R2  jR1 X C  jR2 X C R1R2  j ( R1  R2 ) X C
R2 X C Vi
R2 Vi
=
=
jR1R2  ( R1  R2 ) X C j R1R2 + ( R  R )
1
2
XC
 R2 

 Vi
R1  R2 
R2 Vi

=
=
RR
R1  R2  j 1 2 1 + j  R1R2  1
XC
 R1  R2  X C
R2
V
R1  R2
and Aυ = o =
Vi
 RR 
1 + j  1 2  C
 R1  R2 
or Aυ =
CHAPTER 21

R2 
1


R1  R2 1+ j 2 f ( R1  R2 )C 
297
defining fc =
Aυ =
1
2 ( R1  R2 )C

R2 
1


R1  R2 1+ j f/f c 

R2 
1

  tan 1 f/fc 
R1  R2  1+ ( f/f c )2



R2 
1

|Vi |
with |Vo | =
R1  R2  1 + (f/f c ) 2 


R2
27 k 
for f  fc, Vo =
Vi =
Vi = 0.852Vi
R1  R2
4.7 k  + 27 k 
and Aυ =
at f = fc: Vo = 0.852[0.707]Vi = 0.602Vi
1
1
= 1.02 kHz
fc =

2 ( R1  R2 )C 2 (4.7 k  27 k)0.039  F
b.
c. & d.
20 log10
298
4.7 k  + 27 k 
R1  R2
= 20 log10
27 k 
R2
= 20 log10 1.174 = 1.39 dB
CHAPTER 21
e.
AdB  1.39 dB  0.5 dB = 1.89 dB
AdB = 20 log10 Aυ
1.89 = 20 log10 Aυ
0.0945 = log10 Aυ
V
Aυ = o = 0.80
Vi
f.
θ = tan1 f/fc
39.
R2'  39 k  68 k = 24.79 k
a.
From Section 21.11,
Aυ =
j f/f1
Vo
=
Vi 1  jf/f c
1
1
= 642.01 Hz
=
2 R2 C
2 (24.79 k )(0.01  F)
1
1
=
= 457.47 Hz
fc =
2  R1 + R2  C
2 (10 k  + 24.79 k )(0.01  F)
f1 =
20log10
CHAPTER 21
f
f
 20log10 c
f1
f1
457.5 Hz
= 20 log10
642 Hz
= −2.94 dB
299
b.
f
f
= + tan1 1
f
f1
θ = 45
θ = 54.52
θ = 90  tan1
f = f1:
f = f c:
1
f = f1 = 321 Hz, θ = 63.44
2
1
f=
f1 = 64.2 Hz, θ = 84.29
10
f = 2f1 = 1,284 Hz, θ = 26.57
f = 10f1 = 6420 Hz, θ = 5.71
40.
a.
VTh =
12 k  Vi
= 0.682 Vi
12 k  + 5.6 k 
RTh = 5.6 k  12 k = 3.82 k
f =  Hz: (C  short circuit)
8.2 k  (0.682 Vi )
= 0.465 Vi
Vo =
8.2 k  + 3.82 k 
At f c : V0  0.707(0.465 Vi )  0.329 Vi
300
CHAPTER 21
R2 (0.682 Vi )
0.682 R2 Vi
=
R1  R2  jX C R1  R2  jX C
V
0.682 R2
j 2 f (0.682 R2 )C
and Aυ = o =
=
Vi R1  R2  jX C 1 + j 2 f ( R1  R2 )C
1
1
j f/f1
so that Aυ =
with f1 =
=
1 + j f/f c
2 0.682R2 C 2 0.682(8.2 k )(0.1  F)
= 284.59 Hz
1
1
and fc =
=
2 ( R1  R2 )C 2 (3.82 k  + 8.2 k )(0.1  F)
= 132.41 Hz
voltage-divider rule: Vo =
132.41 Hz
284.59 Hz
= 20 log10 0.465 = 6.65 dB
20 log10 f/f1 = 20 log10
b.
θ = 90  tan1 f/fc = +tan1 fc/f = tan1 132.6 Hz/f
or
CHAPTER 21
301
f
f1
Aυ =
f
1 j
fc
1 j
41.
a.
1
1
= 19.41 kHz

2 R2C 2 (10 k)(820 pF)
1
1
fc =

2 ( R1  R2 )C 2 (10 k  91 k)(820 pF)
= 1.92 kHz
f1 =
20log10
R1  R2
 20log10 10 =  20 dB
R2
b.
θ = tan1 f/f1  tan1 f/fc
f = 10 kHz
10 kHz
10 kHz
 tan1
= 27.25  79.13 = 51.88
θ = tan1
19.41 kHz
1.92 kHz
f = fc: (f1 = 10 fc)
fc
f
θ = tan1
 tan 1 c = tan 1 0.1tan 1 1 = 5.71  45 = 39.29
10 f c
fc
42.
a.
R1 no effect!
Note Section 21.12.
Vo 1 + j ( f/f1 )
=
Vi 1 + j ( f/f c )
1
f1 =
= 2.84 kHz
2 (5.6 k )(0.01  F)
1
fc =
= 904.3 Hz
2 (12 k  + 5.6 k )(0.01  F)
Aυ =
Note Fig. 21.65.
Asymptote at 0 dB from 0  fc
6 dB/octave from fc to f1
12 k  + 5.6 k 


9.95 dB from f1 on  20 log
=  9.95 dB 
5.6
k



302
CHAPTER 21
(b)
Note Fig. 21.67.
From 0 to 26.50 at fc and f1
θ = tan1 f/f1  tan1 f/fc
At f = 1500 Hz (between fc and f1)
θ = tan1 1500 Hz/2.84 kHz  tan1 1500 Hz/904.3 Hz
= 27.83  58.92 = 31.09
43.
a.
V o 1  jf1/f

V i 1  jf c /f
1
1
f1 =
= 945.66 Hz

2 R1C 2 (3.3 k)(0.051  F)
1
1
= 7.59 kHz
fc =
=
2 (R1  R2 )C 2 (3.3 k   0.47 k ) (0.051  F)


0.411 k 
Aυ =
20 log10
R1  R2
3.3 k  + 0.47 k 
=  20 log10 8.02 = 18.08 dB
=  20 log10
R2
0.47 k 
b.
f
f1
+ tan1 c
f
f
f
f
f  f1 :    tan 1 1  tan 1 c
f1
f1
θ = tan1
7.59 kHz
945.66 Hz
 45  82.89  37.89
  tan 1 1  tan 1
945.66 Hz
7.59 kHz
 tan 1
4 kHz
4 kHz
 13.28  62.24  48.96
f  4 kHz:    tan 1
f  f c :    tan 1
f
945.66 Hz
 tan 1 c
7.59 kHz
fc
  tan 1 0.125  tan 1 1
 7.11  45  37.89
CHAPTER 21
303
44.
a.
Note Section 21.13.
1  j ( f1 /f )
Aυ =
1  j (f c /f )
f1 =
fc =
1
1
= 964.58 Hz
=
2 R1C 2 (3.3 k )(0.05  F)
1
1
= 7334.33 Hz
=
2 ( R1  R2 )C 2 (3.3 k   0.5 k ) 0.05  F



0.434 k
Note Fig. 21.72.
20 log10
3.3 k  + 0.5 k 
R1  R2
= 20 log10
= 17.62 dB
0.5 k 
R2
Asymptote at 17.62 dB from 0  f1
+6 dB/octave from f1 to fc
0 dB from fc on
b.
θ = tan1 f1/f + tan1 fc/f
Test at 3 kHz
θ = tan1 964.58 Hz/3.0 kHz + tan1 7334.33 Hz/3.0 kHz
= 17.82 + 67.75 = 49.93  50
Therefore rising above 45 at and near the peak
50 kHz vs 23 kHz  drop about 1 dB at 23 kHz due to 50 kHz break.
Ignore effect of break frequency at 10 Hz.
Assume 2 dB drop at 68 Hz due to break frequency at 45 Hz.
Rough sketch suggests low cut-off frequency of 90 Hz.
Checking: Ignoring upper terms
2
2
 10 Hz 
 45 Hz 
 68 Hz 
A dB  20 log10 1 + 
  20 log10 1 + 
  20 log10 1 + 

 f 
 f 
 f 
= −0.0532 dB − 0.969 dB − 1.96 dB
= −2.98 dB (excellent)
304
2
CHAPTER 21
High frequency cutoff: Try 20 kHz
2
 f 
 f 
A dB =  20log10 1 + 
  20 log10 1 + 

 23 kHz 
 50 kHz 
= −2.445 dB − 0.6445 dB
= −3.09 dB (excellent
2
 BW = 20 kHz  90 Hz = 19,910 Hz  20 kHz
f1 = 90 Hz, f2 = 20 kHz
Testing: f = 100 Hz
10 Hz
45 Hz
68 Hz
f
f
+ tan 1
+ tan 1
 tan 1
 tan 1
23 kHz
50 kHz
f
f
f
= tan1 0.1 + tan1 0.45 + tan1 0.68  tan1 0.00435  tan1 .002
= 5.71 + 24.23 + 34.22  0.249  0.115
= 63.8 vs about 65 on the plot
θ = tan 1
45.
a.
1
A
=
100 Hz 
130 Hz  
f
f 

A max 
1  j
 1  j
 1 + j
1 + j

20 kHz 
50 kHz 
f 
f 

Proximity of 100 Hz to 130 Hz will raise lower cutoff frequency above 130 Hz:
Testing: f = 180 Hz: (with lower terms only)
2
 100 
 130 
A dB = 20 log10 1  
  20 log10 1  

 f 
 f 
2
 100 
 130 
= 20 log10 1  
  20 log10 1  

 180 
 180 
= 1.17 dB  1.82 dB = 2.99 dB  3 dB
CHAPTER 21
2
2
305
Proximity of 50 kHz to 20 kHz will lower high cutoff frequency below 20 kHz:
Testing: f = 18 kHz: (with upper terms only)
2
 f 
 f 
A dB = 20 log10 1  
  20 log10 1  

 20 kHz 
 50 kHz 
2
 18 kHz 
 13 kHz 
= 20 log10 1 + 
  20 log10 1 + 

 20 kHz 
 20 kHz 
= 2.576 dB  0.529 dB = 3.105 dB
2
2
b.
Testing:
f = 1.8 kHz:
100
130
1.8 kHz
1.8 kHz
+ tan 1
 tan 1
 tan 1
1.8 kHz
1.8 kHz
20 kHz
50 kHz
= 3.18 + 4.14  5.14  2.06
= 0.12  0
θ = tan1
47.
flow = fhigh  BW = 36 kHz  35.8 kHz = 0.2 kHz = 200 Hz
Aυ =
306
120

50 
200  
f 

 1  j  1  j
 1  j
f 
f 
36 kHz 

CHAPTER 21
0.05
1
1
+jf
=
=
=
100
100
2000 +jf + 2000
0.05  j
1 j
1 j
f
0.05 f
f
f
+j
2000 and f = 2000 Hz
=
1
f
1+ j
2000
48.
Aυ =
49.
Aυ =
50.
Aυ =
200
1
1


200  j 0.1 f 1  j 0.1 f 1  j f
200
2000
1
f
,
= 1 and f = 2 kHz
A dB = 20 log 20
2
 f  2000
1+ 

 2000 
jf/ 1000
(1 + jf/ 1000)(1 + jf/10,000)
CHAPTER 21
307
51.
f 
f 

1  j
1  j

1000 
2000 
Aυ = 
2
f 

j

1


3000 

2
2
 f 
 f 
AdB = 20 log10 1 +  1  + 20 log10 1 +  2  + 40 log10
 1000 
 2000 
52.
308
1
 f 
1+  3 
 3000 
2
j
2 f
f
f
j
f
= j
= j
= j
,
= j
1000
1000
1000
159.16 Hz 5000
795.78 Hz
2
CHAPTER 21
53.
a.
b.
Woofer − 400 Hz:
XL = 2πfL = 2π(400 Hz)(4.7 mH) = 11.81 
1
1
=
= 10.20 
XC =
2 (400 Hz)(39  F)
2 fC
R  XC = 8  0  10.2090 = 6.3  38.11
( R  X C )(Vi )
(6.3    38.11)(Vi )
=
Vo =
( R  X C )  jX L
(6.3    38.11)  j 11.81 
Vo = 0.673 96.11 Vi
V
and Aυ = o = 0.673 vs desired 0.707 (off by less than 5%)
Vi
Tweeter − 5 kHz:
XL = 2πfL = 2π(5 kHz)(0.39 mH) = 12.25 
1
1
=
= 11.79 
XC =
2 fC 2 ( 5 kHz)(2.7  F)
R  XL = 8  0  12.25  90 = 6.7  33.15
(6.7  33.15)(Vi )
Vo =
(6.7  33.15)  j 11.79 
Vo = 0.678 88.54 Vi
V
and Aυ = o = 0.678 vs 0.707 (off by less than 5%)
Vi
Woofer − 3 kHz:
XL = 2πfL = 2π(3 kHz)(4.7 mH) = 88.59 
1
1
=
= 1.36 
XC =
2 fC 2 ( 3 kHz)(39  F)
R  XC = 8  0  1.36  90 = 1.341  80.35
( R  X C )(Vi )
(1.341    80.35)(Vi )
=
Vo =
( R  X C )  jX L (1.341    80.35)  j 88.59 
Vo = 0.015 170.2 Vi
V
and Aυ = o = 0.015 vs desired 0 (excellent)
Vi
Tweeter − 3 kHz:
XL = 2πfL = 2π(3 kHz)(0.39 mH) = 7.35 
1
1
=
= 19.65 
XC =
2 fC 2 (3 kHz)(2.7  F)
R  XL = 8  0  7.35  90 = 5.42  47.42
( R  X L )(Vi )
(5.42  47.42)( Vi )
=
Vo =
( R  X L )  jX C (5.42  47.42)  j 19.65 
Vo = 0.337 124.24 Vi
V
and Aυ = o = 0.337 (acceptable since relatively close to cut frequency for tweeter)
Vi
CHAPTER 21
309
c.
Mid-range speaker − 3 kHz:

Z = 7.41   22.15



Z = 8.24   33.58



Z = 7.816   37.79
ZVi
(7.816  37.79)Vi
=
= 1.11 8.83 Vi
Z  jX C
7.816  37.79  j1.36 
(7.41    22.15)Vi
ZV1
Vo =
=
= 0.998 46.9 Vi
Z + jX L 7.41    22.15  j 7.35 
V
Aυ = o = 0.998 (excellent)
Vi
V1 =
and
310
CHAPTER 21
Chapter 22
1.
a.
2
(40 mH) 2
M = k L p Ls  L s = M 2 
= 50 mH
(50 mH)(0.8) 2
L pk
b.
ep = N p
es = kNs
c.
ep = L p
es = M
2.
a.
d p
dt
d p
dt
di p
dt
di p
dt
= (20)(0.08 Wb/s) = 1.6 V
= (0.8)(80 t)(0.08 Wb/s) = 5.12 V
= (40 mH)(0.3  103 A/s) = 12 V
= (80 mH)(0.03  103 A/s) = 24 V
k=1
2
b.
(a)
2
( 40 mH )
Ls = M 2 =
= 32 mH
(50 mH)(1) 2
L pk
(b)
ep = 1.6 V, es = kNs
(c)
ep = 15 V, es = 12 V
d p
= (1)(80 t)(0.08 Wb/s) = 6.4 V
dt
k = 0.2
2
3.
(a)
2
(40 mH )
Ls = M 2 =
= 0.8 H
(50 mH)(0.2) 2
L pk
(b)
ep = 1.6 V, es = kNs
(c)
ep = 15 V, es = 12 V
d p
dt
= (0.2)(80 t)(0.08 Wb/s) = 1.28 V
a.
2
( 40 mH ) 2
Ls = M 2 =
= 355.56 mH
(50 mH)(0.3) 2
L pk
b.
ep = N p
es = kNs
c.
d p
dt
d p
dt
= (300 t)(0.08 Wb/s) = 24 V
= (0.9)(25 t)(0.08 Wb/s) = 1.8 V
ep and es the same as problem 1: ep = 15 V, es = 12 V
CHAPTER 22
311
4.
Ns
120 t
Ep =
(40 V) = 240 V
Np
20 t
a.
Es =
b.
Φmax =
a.
Es =
b.
Φm(max) =
6.
Ep =
60 t
Np
Es =
(240 V) = 20 V
720 t
Ns
7.
f=
8.
a.
5.
40 V
Ep
= 7.51 mWb
=
4.4 fN p 4.44(60 Hz)(20 t)
30 t
Ns
(40 V) = 5 V
E p=
240 t
Np
40 V
Ep
=
= 625.63 μWb
4.44 fN p (4.44)(60 Hz)(240 t)
40 V
Ep
= 120 Hz
=
(4.44) N p  m (max) (4.44)(20 t)(3.75 mWb)
1
IL = aIp =   (2 A) = 0.4 A
5
2 
VL = ILZL =  A  (2 ) = 0.8 V
5 
2
b.
9.
10.
11.
Zp =
1
Zin = a2ZL =   2  = 0.08 
5
Vg
Ip
=
120 V
= 30 
4A
1
Vg = aVL =   (600 V) = 150 V
4
V g 150 V
Ip =
=
= 37.5 A
4
Zi
IL = Is =
VL 240 V
= 12 A
=
Z L 20 
I s = a = N p  12 A = N p
0.05 A 50
Ip
Ns
50(12)
Np =
= 12,000 turns
0.05
312
CHAPTER 22
12.
a.
a=
1
N p 400 t
=
=
1200
t
3
Ns
2
1
Zi = a ZL =   [12  + j12 ] = 1.333  + j1.333  = 1.885  45
3
Ip = Vg/Zi = 100 V/1.885 Ω = 53.05 A
2
13.
IL = aIp =
a.
Zp = a2ZL  a =
b.
Zp
ZL
Zp =
10 V
Vp
=
= 36 
I p 20 V/ 72 
a=
36 
=3
4
1
1
1
Vs Ns 1
=
=  V s = V p = (10 V) = 3 V
3
3
3
Vp Np 3
P=
14.
1
(53.05 A) = 17.68 A, VL = ILZL = (17.68 A)(16.97 Ω) = 300 V
3
b.
Vs2 (3.33 V ) 2
= 2.78 W
=
Zs
4
a.
Re = Rp + a2Rs = 4  + (4)2 1  = 20 
b.
Xe = Xp + a2Xs = 12  + (4)2 2  = 44 
c.
d.
Ip =
Vg
Zp
=
120 V 0
120 V 0
=
= 0.554 A 11.73
20  + 192  + j 44  212  + j 44 
2
e.
aVL =
or
f.

g.
VL =
CHAPTER 22
a R LV g
= Ipa2RL
2
( R e  a R L)  j X e
VL = aIpRL0 = (4)(0.554 A 11.73)(20  0) = 26.59 V 11.73
1
Ns
Vg = (120 V) = 30 V
4
Np
313
15.
a.
b.
c.
4t
=4
Ns 1t
Re = Rp + a2Rs = 4  + (4)2 1  = 20 
Xe = Xp + a2Xs = 12  + (4)2 2  = 44 
Zp = Z Re + Z X e + a 2 Z X L = 20  + j44  + j(4)2 20 
= 20  + j44  + j320  = 20  + j364  = 364.55  86.86
a=
Ip =
Np
Vg
=
=
Zp
120 V 0
= 329.17 mA 86.86
364.55  86.86
VRe = (Iθ)(Re0) = (329.17 mA 86.86)(20  0)
= 6.58 V 86.86
VX e = (Iθ)(Xe90) = (329.17 mA 86.86)(44  90)
= 14.48 V 3.14
VX L = I(a2 Z X L ) = (329.17 mA 86.86)(320 Ω 90)
= 105.33 V 3.14
16.
a.
a = Np/Ns = 4 t/1 t = 4, Re = Rp + a2Rs = 4  + (4)2 1  = 20 
Xe = Xp + a2Xs = 12  + (4)2 2  = 44 
Zp = Re + jXe  ja2XC = 20  + j44   j(4)2 20 
= 20  + j44   j320  = 20   j276  = 276.72 Ω 85.86
b.
Ip =
c.
Vg
Zp
=
120 V 0
= 0.43 A 85.86
276.72    85.86
V Re = (Ipθ)(Re0) = (0.43 A 85.86)(20  0) = 8.6 V 85.86
V X e = (Ipθ)(Xe90) = (0.43 A 85.86)(44  90) = 18.92 V 175.86
V X C = (Ipθ)(a XC90) = (0.43 A 85.86)(320  90) = 137.60 V 4.14
2
17.

18.
Coil 1:
L1  M12
Coil 2:
L2  M12
LT = L1 + L2  2M12 = 4 H + 7 H  2(1 H) = 9 H
19.
L T (  )  L1  L2  2M 12
M12 = k L1 L2 = (0.8) (200 mH)(600 mH) = 277 mH
LT (+) = 200 mH + 600 mH + 2(277 mH) = 1.35 H
314
CHAPTER 22
20.
M23 = k L2 L3  1 (1 H)(4 H) = 2 H
L1 + M12  M13 = 2 H + 0.2 H  0.1 H = 2.1 H
L2 + M12  M23 = 1 H + 0.2 H  2 H = 0.8 H
L3  M23  M13 = 4 H  2 H  0.1 H = 1.9 H
LT = 2.1 H  0.8 H + 1.9 H = 3.2 H
Coil 1:
Coil 2:
Coil 3:
21.
E1  I1[ Z R1 + Z L1 ]  I2[Zm] = 0
I2[ Z L2 + Z RL ] + I1[Zm] = 0
──────────────────────
I1( Z R1 + Z L1 ) + I2(Zm) = E1
I1(Zm) + I2( Z L2 + Z RL ) = 0
───────────────────────
22.
Zi = Zp +
Xm = ωM 90
( M ) 2
( M ) 2
= Rp + j X Lp +
Zs + ZL
Rs + jX Ls + RL
Rp = 2 , X L p = ωLp = (103 rad/s)(8 H) = 8 k
Rs = 1 , X Ls = ωLs = (103 rad/s)(2 H) = 2 k
M = k L p L s = 0.05 (8 H)(2 H) = 0.2 H
Zi = 2  + j8 k +
(103 rad/s  0.2 H ) 2
1  + j 2 k  + 20 
4  104 
21  j 2  103
= 2  + j8 k + 0.21   j19.99  = 2.21  + j7980 
Zi = 7980  89.98
= 2  + j8 k +
23.
Np
Vp
2400 V
= 20
120 V
a.
a=
b.
10,000 VA = VsIs  Is =
c.
Ip =
d.
a=
Is =
CHAPTER 22
Ns
=
Vs
=
10,000 VA 10, 000 VA
= 83.33 A
=
120 V
Vs
10,000 VA 10,000 VA
=
= 4.17 A
2400 V
Vp
Vp
Vs
=
120 V
1
= 0.05 =
2400 V
20
10,000 VA
= 4.17 A, Ip = 83.33 A
2400 V
315
24.
Is = I1 = 2 A, Ep = VL = 40 V
Es = Vs  VL = 200 V  40 V = 160 V
200 V
(2 A) = 10 A
VgI1 = VLIL  IL = Vg/VL  I1 =
40 V
Ip + I1 = IL  Ip = IL  I1 = 10 A  2A = 8 A
25.
a.
Es =
Ns
Ep
Np
25 t
(100 V 0) = 25 V 0 = VL
100 t
E
25 V 0
= 5 A 0 = IL
Is = s =
Z L 5  0
=
2
26.
b.
N 
 100 t 
2
Zi = a ZL =  p  Z L  
 5  0 = (4) 5  0 = 80  0
 25 t 
 Ns 
c.
1
1
Z1/ 2 = Zi = (80  0) = 20  0
4
4
a.
b.
2
2
15 t
N2
E1 =
(60 V 0) = 10 V 0
90 t
N1
45 t
E3 = N 3 E1 =
(60 V 0) = 30 V 0
90 t
N1
10 V 0
I2 = E 2 =
= 1.25 A 0
Z 2 8  0
30 V 0
I3 = E 3 =
= 6 A 0
Z3 5  0
E2 =
1
=
R1
1
1

2
( N1 / N 2 ) R2 ( N1 / N 3 ) 2R3
1
1

2
(90 t /15 t ) 8  (90 t / 45 t ) 2 5 
1
1
1
= 0.05347 S


R1 288  20 
R1 = 18.70 
=
27.
a.
N2
 40 t 
E1 = 
 (120 V 60) = 40 V 60
N1
 120 t 
E
40 V 60
= 3.33 A 60
I2 = 2 =
Z 2 12  0
E2 =
N3
 30 t 
E1 = 
 (120 V 60) = 30 V 60
N1
 120 t 
E
30 V 60
I3 = 3 =
= 3 A 60
Z3 10  0
E3 =
316
CHAPTER 22
b.
1
=
R1
1
1

2
( N1 / N 2 ) R2 ( N1 / N 3 ) 2R3
1
1

2
(120 t / 40 t ) 12  (120 t / 30 t ) 210 
1
1
1

=
= 0.0155 S
108

160

R1
1
= 64.52 
R1 =
0.0155 S
=
28.
ZM = Z M12 = ωM12 90
E  I1Z1  I1 Z L1  I1(Zm)  I2(+Zm)  I1 Z L2 + I2 Z L2  I1(Zm) = 0
E  I1(Z1 + Z L1  Zm + Z L2  Zm)  I2(Zm  Z L2 ) = 0
or
I1(Z1 + Z L1 + Z L2  2 Zm) + I2(Zm  Z L2 ) = E
──────────────────────────────────────────────
I2Z2  Z L2 (I2  I1)  I1(+Zm) = 0
or
I1(Zm  Z L2 ) + I2(Z2 + Z L2 ) = 0
──────────────────────────
E1  I1Z1  I1 Z L1  I2( Z M12 )  I3(+ Z M13 ) = 0
29.
or
E1  I1[Z1 + Z L1 ] + I2 Z M12  I3 Z M13 = 0
────────────────────────────────
I2(Z2 + Z3 + Z L2 ) + I3Z2  I1( Z M12 ) = 0
or
I2(Z2 + Z3 + Z L2 ) + I3Z2 + I1 Z M12 = 0
────────────────────────────────
I3(Z2 + Z4 + Z L3 ) + I2Z2  I1(+ Z M13 ) = 0
or
I3(Z2 + Z4 + Z L3 ) + I2Z2  I1 Z M13 = 0
─────────────────────────────
Z M12 I2 +
[Z1 + Z L1 ]I1 
Z M13 I3 = E1
Z M12 I1  [Z2 + Z3 + Z L2 ]I2 +
Z2I3 = 0

Z M13 I1
Z2I2 + [Z2 + Z4 + Z L3 ]I3 = 0
────────────────────────────────────────
CHAPTER 22
317
Chapter 23
1.
2.
a.
E = EL/ 3 = 208 V/1.732 = 120.1 V
b.
V = E = 120.1 V
c.
I =
d.
IL = I = 12.01 A
a.
E = EL/ 3 = 208 V/1.732 = 120.1 V
b.
V = E = 120.1 V
c.
Z = 12   j16 Ω = 20  53.13
d.
IL = I = 6 A
b.
V = 120.1 V
I =
3.
4.
V  120.1 V
 6A
=
20 
Z
a.
E = 120.1 V
c.
Z = (10  0  (10  90) = 7.071  45
120.1 V
V
I =  =
= 16.98 A
Z  7.071 
d.
IL = 16.98 A
a.
θ2 = 120, θ3 = 120
b.
Van = 120 V 0, Vbn = 120 V 120, Vcn = 120 V 120
c.
5.
V  120.1 V
= 12.01 A
=
10 
R
120 V 0
Ian = V an =
= 6 A 0
Z an 20  0
120 V   120
Ibn = V bn =
= 6 A 120
20  0
Z bn
120 V 120
Icn = V cn =
= 6 A 120
20  0
Z cn
d.
IL = I = 6A
a.
θ2 = 120, θ3 = +120
b.
Van = 120 V 0, Vbn = 120 V 120, Vcn = 120 V 120
c.
Z = 9  + j12  = 15  53.13
e.
VL =
3 V =
3 (120 V) = 207.8 V
120 V 0
120 V   120
= 8 A 53.13, Ibn =
= 8 A 173.13
15  53.13
15  53.13
120 V 120
= 8 A 66.87
Icn =
15  53.13
Ian =
318
CHAPTER 23
e.
6.
a, b.
c.
IL = I  = 8 A
f.
EL =
3 E = (1.732)(120 V) = 207.85 V
The same as problem 4.
Z = 6  0  8  90 = 4.8  36.87
120 V 0
= 25 A 36.87
Ian = V an =
Z an 4.8    36.87
120 V   120
Ibn = V bn =
= 25 A 83.13
Z bn 4.8    36.87
120 V 120
Icn = V cn =
= 25 A 156.87
Z cn 4.8    36.87
d.
7.
IL = I = 25 A
e.
VL
VL =
3 V =
3 (120 V) = 207.84 V
220 V
= 127.0 V
3 1.732
Z = 10   j10  = 14.42  45
V = Van = Vbn = Vcn =
=
127 V
V
=
= 8.98 A
Z  14.142 
IL = IAa = IBb = ICc = I = 8.98 A
I = Ian = Ibn = Icn =
8.
Z = 12  + j16  = 20 53.13
V  50 V
=
= 2.5 A
Z  20 
ZT  = 13  + j16  = 20.62  50.91
I =
V = I Z T  = (2.5 A)(20.62 ) = 51.55 V
VL =
9.
a.
3 V =
 3  (51.55 V) = 89.29 V
22 kV
  30 = 12.7 kV 30
3
22 kV
EBN =
  150 = 12.7 kV 150
3
22 kV
ECN =
90 = 12.7 kV 90
3
EAN =
CHAPTER 23
319
b, c. IAa = Ian =
E AN
12.7 kV   30
=
(30  + j 40 ) + (0.4 k  + j1 k )
Z AN
12.7 kV   30
12.7 kV   30
=
430  + j1040  1125.39  67.54
= 11.29 A 97.54
12.7 kV   150
= 11.29 A 217.54
IBb = Ibn = E BN =
Z BN 1125.39  67.54
12.7 kV 90
ICc = Icn = E CN =
= 11.29 A 22.46
Z CN 1125.39  67.54
=
10.
11.
12.
13.
320
d.
Van = IanZan = (11.29 A 97.54)(400 + j1000)
= (11.29 A 97.54)(1077.03  68.2)
= 12.16 kV 29.34
Vbn = IbnZbn = (11.29 A 217.54)(1077.03 68.2)
= 12.16 kV 149.34
Vcn = IcnZcn = (11.29 A 22.46)(1077.03 68.2)
= 12.16 kV 90.66
a.
E = EL/ 3 = 208 V/1.732 = 120.1 V
c.
I =
a.
E = EL/ 3 = 208 V/1.732 = 120.1 V
c.
Z = 6.8  + j14  = 15.564  64.09
208 V
V
I =  =
= 13.36 A
Z  15.564 
d.
IL = 3 I = (1.732)(13.36 A) = 23.14 A
V  208 V
= 10.4 A
=
Z  20 
b.
V = EL = 208 V
d.
IL = 3 I = (1.732)(10.4 A) = 18 A
b.
V = EL = 208 V
b.
V = 208 V
Z = 18  0  18  90 = 12.728  45
a.
E = VL/ 3 = 208 V/ 3 = 120.09 V
c.
I =
d.
IL = 3 I = (1.732)(16.34 A) = 28.30 A
a.
θ2 = 120, θ3 = +120
b.
Vab = 208 V 0, Vbc = 208 V 120, Vca = 208 V 120
208 V
V
= 16.34 A
=
Z  12.728 
CHAPTER 23
c.
d.
14.

208 V 0
Iab = V ab =
= 9.46 A 0
Z ab 22  0
208 V 120
Ibc = V bc =
= 9.46 A 120
22  0
Z bc
208 V 120
V
= 9.46 A 120
Ica = ca =
22  0
Z ca
e.
IL = 3 I = (1.732)(9.46 A) = 16.38 A
f.
E = EL/ 3 = 208 V/1.732 = 120.1 V
a.
θ2 = 120, θ3 = +120
b.
Vab = 208 V 0, Vbc = 208 V 120, Vca = 208 V 120
c.

d.
Z = 100   j100  = 141.42 45
208 V 0
Iab = V ab =
= 1.47 A 45
141.42
   45
Z ab
208 V   120
Ibc = V bc =
= 1.47 A 75
Z bc 141.42    45
208 V 120
Ica = V ca =
= 1.47 A 165
Z ca 141.42    45
e.
IL = 3 I = (1.732)(1.471 A) = 2.55 A
f.
E = EL/ 3 = 208 V/1.732 = 120.1 V
15. a, b.
The same as problem 13.
c.

d.
Z = 3  0  4  90 = 2.4  36.87
208 V 0
= 86.67 A 36.87
Iab = V ab =
Z ab 2.4  36.87
208 V   120
Ibc = V bc =
= 86.67 A 156.87
Z bc 2.4  36.87
208 V 120
Ica = V ca =
= 86.67 A 83.13
Z ca 2.4  36.87
CHAPTER 23
321
e.
IL =
3 I = (1.732)(86.67 A) = 150.11 A
16.
Vab = Vbc = Vca = 220 V
Z = 10  + j10  = 14.142 45
220 V
V
= 15.56 A
Iab = Ibc = Ica =  =
14.142

Z
17.
a.
f.
E = 120.1 V
16 kV 0
16 kV 0
=
Iab = V ab =
Z ab 300  + j1000  1044.03  73.30
Iab = 15.33 A 73.30
16 kV   120
= 15.33 A 193.30
Ibc = V bc =
Z bc 1044.03  73.30
16 kV 120
Ica = V ca =
= 15.33 A 46.7
Z ca 1044.03  73.30
b.
IAa  Iab + Ica = 0
IAa = Iab  Ica = 15.33 A 73.30  15.33 A 46.7
= (4.41 A  j14.68 A)  (10.51 A + j11.16 A)
= 4.41 A  10.51 A  j(14.68 A + 11.16 A)
= 6.11 A  j25.84 A = 26.55 A 103.30
IBb + Iab = Ibc
IBb = Ibc  Iab = 15.33 A 193.30  15.33 A 73.30
= 26.55 A 136.70
ICc + Ibc = Ica
ICc = Ica  Ibc = 15.33 A 46.7  15.33 A 193.30
= 26.55 A 16.70
c.
EAB = IAa(10  + j20 ) + Vab  IBb(22.361  63.43)
= (26.55 A 103.30)(22.361  63.43) + 16 kV 0
 (26.55 A 136.70)(22.361  63.43)
= (455.65 V  j380.58 V) + 16,000 V  (557.42 V  j204.32 V)
= 17.01 kV  j176.26 V
= 17.01 kV 0.59
EBC = IBb(22.361  63.43) + Vbc  ICc(22.361  63.53)
= (26.55 A 136.70)(22.361  63.53) + 16 kV 120
 (26.55 A 16.70)(22.361  63.53)
= 17.01 kV 120.59
ECA = ICc(22.361  63.43) + Vca  IAa(22.361  63.43)
= 17.01 kV 119.41
322
CHAPTER 23
18.
19.
a.
E = EL = 208 V
c.
I =
a.
E = EL = 208 V
c.
I =
20. a, b.
c.
21.
22.
V  120.09 V
=
= 7.08 A
Z  16.971 
208 V
V = E L =
= 120.1 V
3 1.732
d.
IL = I   4 A
b.
V = EL 3 = 120.09 V
d.
IL = I = 7.08 A
The same as problem 18.
Z = 15  0  20  90 = 12  36.87
I =
d.
V  120.1 V
=
= 4.00 A
30 
Z
b.
V  120.1 V
=
 10 A
12 
Z
IL = I  10 A
120 V 120 V
=
= 69.28 V
1.732
3
69.28 V
Ian = Ibn = Icn =
= 2.89 A
24 
IAa = IBb = ICc = 2.89 A
Van = Vbn = Vcn =
120 V
= 69.28 V
3
Z = 10  + j20  = 22.36 63.43
Van = Vbn = Vcn =
Ian = Ibn = Icn =
V
Z
=
69.28 V
= 3.10 A
22.36 
IAa = IBb = ICc = I = 3.10 A
23.
Van = Vbn = Vcn = 69.28 V
Z = 20  0  15  90 = 12  53.13
69.28 V
= 5.77 A
12 
IAa = IBb = ICc = 5.77 A
Ian = Ibn = Icn =
24.
a.
E = EL = 440 V
c.
I =
CHAPTER 23
V  440 V
=2A
=
Z  220 
b.
V = EL = E = 440 V
d.
IL = 3 I = (1.732)(2 A) = 3.46 A
323
25.
a.
E = EL = 440 V
c.
Z = 12   j9  = 15  36.87
d.
26. a, b.
c.
I =
V  440 V
= 29.33 A
=
Z  15 
IL =
3 I = (1.732)(29.33 A) = 50.8 A
Z = 22  0  22  90 = 15.56  45
440 V
V
= 28.28 A
=
Z  15.56 
d.
IL =
a.
θ2 = 120, θ3 = +120
b.
Vab = 100 V 0, Vbc = 100 V 120, Vca = 100 V 120
c.

d.
28.
V = EL = 440 V
The same as problem 24.
I =
27.
b.
3 I = (1.732)(28.28 A) = 48.98 A
V ab 100 V 0
=
= 5 A 0
Z ab 20  0
100 V   120
Ibc = V bc =
= 5 A 120
20  0
Z bc
100 V 120
Ica = V ca =
= 5 A 120
20  0
Z ca
Iab =
e.
IAa = IBb = ICc =
3 (5 A) = 8.66 A
a.
θ2 = 120, θ3 = +120
b.
Vab = 100 V 0, Vbc = 100 V 120, Vca = 100 V 120
c.

d.
Z = 12  + j16  = 20  53.13
100 V 0
V ab
=
= 5 A 53.13
Z ab 20  53.13
100 V   120
Ibc = V bc =
= 5 A 173.13
Z bc 20  53.13
Iab =
324
CHAPTER 23
100 V 120
Ica = V ca =
= 5 A 66.87
Z ca 20  53.13
29.
e.
IAa = IBb = ICc =
a.
θ2 = 120, θ3 = 120
b.
Vab = 100 V 0, Vbc = 100 V 120, Vca = 100 V 120
c.

d.
Z = 20  0  20  90 = 14.14  45
3 I = (1.732)(5 A) = 8.66 A
100 V 0
= 7.07 A 45
14.14    45
100 V   120
Ibc =
= 7.07 A 75
14.14    45
100 V 120
= 7.07 A 165
Ica =
14.14    45
Iab =
e.
30.
IAa = IBb = ICc =
 3  (7.07 A) = 12.25 A
PT = 3I2 R = 3(6 A)2 12  = 1296 W
QT = 3I2 X  = 3(6 A)2 16  = 1728 VAR(C)
ST =
PT  QT = 2160 VA
2
2
1296 W
= 0.6 (leading)
Fp = PT =
2160
VA
ST
31.
V = 120 V, I = 120 V/20  = 6 A
PT = 3I2 R = 3(6 A)2 20  = 2160 W
QT = 0 VAR
ST = PT = 2160 VA
2160 W
Fp = PT =
=1
S T 2160 VA
32.
PT = 3I2 R = 3(8.98 A)2 10  = 2419.21 W
QT = 3I2 X  = 3(8.98 A)2 10  = 2419.21 VAR(C)
ST =
PT2  Q T2 = 3421.28 VA
Fp =
PT
2419.21 W
= 0.7071 (leading)
=
ST 3421.28 VA
CHAPTER 23
325
33.
V = 208 V
V 2 
(208 V ) 2
PT = 3    = 3 
= 7210.67 W
18 
 R 
V2 
(208 V ) 2
= 7210.67 VAR(C)
QT = 3    = 3 
18 
 X 
PT  QT = 10,197.42 VA
2
ST =
2
7210.67 W
= 0.707 (leading)
Fp = PT =
10,197.42
VA
ST
34.
PT = 3I2 R = 3(1.471 A)2 100  = 649.15 W
QT = 3I2 X  = 3(1.471 A)2 100  = 649.15 VAR(C)
35.
ST =
PT2  QT2 = 918.04 VA
Fp =
PT
649.15 W
= 0.7071 (leading)
=
ST 918.04 VA
PT = 3I2 R = 3(15.56 A)2 10  = 7.26 kW
QT = 3I2 X  = 3(15.56 A)2 10  = 7.26 kVAR
36.
ST =
PT2  Q T2 = 10.27 kVA
Fp =
PT
7.263 kW
= 0.7071 (lagging)
=
ST 10.272 kVA
PT = 3
2
2
2
2
V  3(120.1 V )
=
= 2884.80 W
15 
R
QT = 3
V  3(120.1 V )
=
= 2163.60 VAR(C)
20 
X
PT  QT = 3605.97 VA
P
2884.80 W
= 0.8 (leading)
Fp = T =
ST 3605.97 VA
ST =
37.
2
2
Z = 10  + j20  = 22.36  63.43
V L 120 V
= 69.28 V
=
3 1.732
69.28 V
V
I =  =
= 3.098 A
Z  22.36 
V =
PT = 3I2 R = 3(3.098 A)2 10  = 287.93 W
326
CHAPTER 23
QT = 3I2 X  = 3(3.098 A)2 20 Ω = 575.86 VAR
ST =
PT2  QT2 = 643.83 VA
Fp =
PT
287.93 W
= 0.447 (lagging)
=
ST 643.83 VA
2
38.
39.
2
V  3(440 V )
=
= 26.4 kW
22 
R
QT = PT = 26.4 kVAR(L)
PT = 3
ST =
PT2  QT2 = 37.34 kVA
Fp =
PT
26.4 kW
= 0.707 (lagging)
=
ST 37.34 kVA
Z = 12  + j16  = 20  53.13
I =
V  100 V
=5A
=
Z  20 
PT = 3I2 R = 3(5 A)2 12  = 900 W
QT = 3I2 X  = 3(5 A)2 16  = 1200 VAR(L)
2
2
PT  QT = 1500 VA
P
900 W
Fp = T =
= 0.6 (lagging)
ST 1500 VA
ST =
40.
PT = 3 ELIL cos θ
4800 W = (1.732)(200 V)IL (0.8)
IL = 17.32 A
I
17.32 A
I = L =
= 10 A
1.732
3
θ = cos1 0.8 = 36.87
200 V 0
V
Z =  =
= 20  36.87 = 16  + j12 
I 10 A   36.87
41.
PT =
3 ELIL cos θ
1200 W = 3 (208 V)IL(0.6)  IL = 5.55 A
V
208 V
V = L =
= 120.1 V
3 1.732
θ = cos1 0.6 = 53.13 (leading)
120.1 V 0
V
Z =  =
= 21.64  53.13 = 12.98
  j 17.31







I 5.55 A 53.13
R
XC
CHAPTER 23
327
42.
Δ:
Z = 15  + j20  = 25 53.13
V
I =
Z
=
125 V
=5A
25 
PT = 3I2 R = 3(5 A)2 15  = 1125 W
QT = 3I2 X  = 3(5 A)2 20  = 1500 VAR(L)
Y:
V = VL/ 3 = 125 V/1.732 = 72.17 V
Z = 3   j4  = 5 53.13
I =
V  72.17 V
= 14.43 A
=
5
Z
PT = 3I2 R = 3(14.43 A)2 3  = 1874.02 W
QT = 3I2 X  = 3(14.43 A)2 4  = 2498.7 VAR
PT = 1125 W + 1874.02 W = 2999.02 W
QT = 1500 VAR(L)  2498.7 VAR(C) = 998.7 VAR(C)
43.
ST =
PT2  QT2 = 3161 VA
Fp =
PT 2999.02 W
= 0.949 (leading)
=
3161 VA
ST
a.
E =
c.
d.
e.
16 kV
= 9,237.6 V
3
b.
IL = I = 80 A
1200 kW
= 400 kW
3
P4Ω = (80 A)24  = 25.6 kW
PT = 3P = 3(25.6 kW + 400 kW) = 1276.8 kW
P L =
Fp =
PT
, ST =
ST
Fp =
1, 276.8 kW
= 0.576 lagging
2,217.025 kVA
3 VLIL =
3 (16 kV)(80 A) = 2,217.025 kVA
θL = cos1 0.576 = 54.83 (lagging)
E AN0  80A   54.83°
IAa =

Z T 54.83

given

for entire load
328
CHAPTER 23
44.
f.
Van = EAN  IAa(4  + j20 )
= 9237.6 V 0  (80 A 54.83)(20.396  78.69)
= 9237.6 V 0  1631.68 V 23.86
= 9237.6 V  (1492.22 V + j660 V)
= 7745.38 V  j660 V
= 7773.45 V 4.87
g.
Z =
h.
Fp(entire system) = 0.576 (lagging)
Fp(load) = 0.643 (lagging)
V an 7773.45 V   4.87
=
= 97.168  49.95
80 A   54.83
I Aa
= 62.52
 + j 74.38







R
XC
i.

1276.8 kW  3(25.6 kW)
η = P o = P i P lost =
= 0.9398  93.98%
1276.8 kW
Pi
Pi
a.

b.
V =
220 V
= 127.02 V, Z = 10   j10  = 14.14 45
3
127.02 V
V
I =  =
= 8.98 A
Z  14.14 
PT = 3I2 R = 3(8.98 A)2 10  = 2419.2 W
Each wattmeter:
2419.2 W
= 806.4 W
3
45.
b.
PT = 5899.64 W, Pmeter = 1966.55 W
46.
a.

b.
PT = P + Ph = 85 W + 200 W = 285 W
c.
0.2  P  = 0.5
Ph
100 W
Ph = P  =
= 200 W
0.5
0.5
PT = Ph  P = 200 W  100 W = 100 W
CHAPTER 23
329
48.
a.
208 V 0
Iab = E AB =
= 20.8 A 0
R0 10  0
208 V   120 208 V   120
E BC
= 14.708 A 165
Ibc =
=

R  jX L
10   j10 
14.142   45
208 V 120
208 V 120
E CA
Ica =
= 14.708 A 165
=

R  jX C 10   j10  14.142    45
b.
IAa + Ica  Iab = 0
IAa = Iab  Ica
= 20.8 A 0  14.708 A 165
= 20.8 A  (14.207A + j3.807 A)
= 35.007 A  j3.807 A
= 35.213 A 6.207
IBb + Iab  Ibc = 0
IBb = Ibc  Iab
= 14.708 A 165  20.8 A 0
= (14.207 A  j3.807 A)  20.8 A
= 35.007 A  j3.807 A
= 35.213 A 173.79
ICc + Ibc  Ica = 0
ICc = Ica  Ibc
= 14.708 A 165  14.708 A 165
= (14.207 A + j3.807 A)  (14.207 A  j3.807 A)
= 7.614 A 90
c.
P1 = VacIAa cos IVAaca Vca = Vca  θ  180 = 208 V 120  180
= 208 V 60
IAa = 35.213 A 6.207
P1 = (208 V)(35.213 A) cos 53.793
= 4.326 kW
P2 = VbcIBb cos IVBbbc Vbc = 208 V 120
IBb = 35.213 A 173.79
P2 = (208 V)(35.213 A) cos 53.79
= 4.327 kW
d.
330
PT = P1 + P2 = 4.326 kW + 4.327 kW = 8.653 kW
CHAPTER 23
49.
a.
b.
c.
V = E = E L = 120.09 V
3
120.09 V
Ian = V an =
= 8.49 A
Z an 14.142 
120.09 V
V
= 7.08 A
Ibn = bn =
Z bn 16.971 
120.09 V
Icn = V cn =
= 42.47 A
Z cn 2.828 
2
2
PT = I an
10  + I bn
12  + I cn2 2 
= (8.49 A)2 10  + (7.08 A)2 12  + (42.47 A)2 2 
= 720.80 W + 601.52 W + 3.61 kW
= 4.93 kW
QT = PT = 4.93 kVAR(L)
50.
ST =
PT2  QT2 = 6.97 kVA
Fp =
PT
= 0.707 (lagging)
ST
d.
Ean = 120.09 V30, Ebn = 120.09 V150, Ecn = 120.09 V90
120.09 V   30 120.09 V   30
=
= 8.49 A 75
Ian = E an =
10  + j10 
14.142  45
Z an
120.09 V   150 120.09 V   150
Ibn = Ebn =
=
= 7.08 A 195
12  + j12 
16.971  45
Z bn
120.09 V 90 120.09 V 90
Icn = Ecn =
=
= 42.47 A 45
2  + j2 
2.828  45
Z cn
e.
IN = Ian + Ibn + Icn
= 8.49 A 75 + 7.08 A 195 + 42.47 A45
= (2.02 A  j8.20 A) + (6.84 A + j1.83 A) + (30.30 A + j30.30 A)
= 25.66 A  j23.93 A
= 35.09 A 43.00
Z1 = 12   j16  = 20  53.13, Z2 = 3  + j4  = 5  53.13
Z3 = 20  0
EAB = 200 V0, EBC = 200 V 120, ECA = 200 V 120
ZΔ = Z1Z2 + Z1Z3 + Z2Z3
= (20  53.13)(5  53.13) + (20  53.13)(20  0)
+ (5  53.13)(20  0)
= 100  0 + 400  53.13 + 100  53.13
= 100  + (240   j320 ) + (60  + j80 )
= 400   j240 
= 466.48  30.96
CHAPTER 23
331
Ian =
E AB Z3  ECA Z 2 (200 V 0)(20  0)  (200 V 120)(5  53.13)
=
Z
Z
4000 A 0  1000 A 173.13
= 10.71 A 29.59
466.48   30.96
E Z  E AB Z3 (200 V   120)(20    53.13)  (200 V 0)(20  0)
Ibn = BC 1
=
Z
Z
=
4000 A   173.13  4000 A 0
= 17.12 A 145.61
466.48   30.96
E Z  E BC Z1 (200 V 120)(5 53.13)  (200 V   120)(20   53.13)
Icn = CA 2
=
Z
Z
=
1000 A 173.13  4000 A   173.13
= 6.51 A 42.32
466.48   30.96
2
2
PT = I an
12  + I bn
4  + I cn2 20 
= 1376.45 W + 1172.38 W + 847.60 W = 3396.43 W
2
2
QT = I an
16  + I bn
3  = 1835.27 VAR(C) + 879.28 VAR(L) = 955.99 VAR(C)
=
332
ST =
PT2  QT2 = 3508.40 VA
Fp =
PT
3396.43 W
= 0.968 (leading)
=
ST 3508.40 VA
CHAPTER 23
Chapter 24
1.
a.
positive-going
d.
Amplitude = 8 V  2 V = 6 V
e.
f.
g.
2.
Vb = 2 V
tp = 0.2 ms
V1 V 2
 100%
V
8V + 7.5 V
V=
= 7.75 V
2
8 V  7.5 V
% tilt =
 100% = 6.5%
7.75 V
1
1
1
prf = =

= 625 kHz
T (2.0 ms  0.4 ms) 1.6 ms
tp
T
 100% 
0.2 ms
 100% = 12.5%
1.6 ms
negative-going
d.
8 mV ( from base line level)
b.
+7 mV
c.
3 μs
8 mV  7 mV 15 mV
=
= 7.5 mV
2
2

8 mV  (7 mV)
% Tilt = V 1 V 2  100% =
 100%
V
7.5 mV
1 mV
=
 100% = 13.3%
7.5 mV
V=
f.
T = 15 μs  7 μs = 8 μs
1
1
= 125 kHz
prf = =
T 8 s
g.
Duty cycle =
a.
positive-going
d.
Amplitude = (30  10)mV = 20 mV
e.
c.
% tilt =
a.
e.
3.
b.
3 s
tp
 100% =
 100% = 37.5%
T
8 s
b.
Vb = 10 mV
c.
8
tp =   4 ms = 3.2 ms
 10 

% tilt = V 1 V 2  100%
V
30 mV + 28 mV
V=
= 29 mV
2
30 mV  28 mV
% tilt =
 100%  6.9%
29 mV
CHAPTER 24
333
4.
tr  (0.2 div.)(2 ms/div.) = 0.4 ms
tf  (0.4 div.)(2 ms/div.) = 0.8 ms
5.
tilt =
V1  V2
V V
= 0.1 with V = 1 2
V
2
Substituting V into top equation,
V1  V2
0.95 V1
= 0.1 leading to V2 =
or V2 = 0.905(15 mV) = 13.58 Mv
V1  V2
1.05
2
6.
7.
a.
tr = 80% of straight line segment
= 0.8(2 μs) = 1.6 μs
b.
tf = 80% of 4 μs interval
= 0.8(4 μs) = 3.2 μs
c.
At 50% level (10 mV)
tp = (8  1)μs = 7 μs
d.
prf =
a.
T = (4.8  2.4)div. 50  s/div. = 120 μs
c.
8.
334
1
1
=
= 50 kHz
T 20  s
b.
f=
1
1

= 8.33 kHz
T 120 s
Maximum Amplitude: (2.2 div.)(0.2 V/div.) = 0.44 V = 440 mV
Minimum Amplitude: (0.4 div.)(0.2 V/div.) = 0.08 V = 80 mV
T = (3.6  2.0)ms = 1.6 ms
1
1
= 625 Hz
prf = =
T 1.6 ms
tp
0.2 ms
Duty cycle =  100% =
 100% = 12.5%
T
1.6 ms
CHAPTER 24
9.
T = (15  7)μs = 8 μs
1
1
= 125 kHz
prf = 
T 8 s
tp
(20  15)  s
5
 100% =  100% = 62.5%
Duty cycle =  100% 
T
8 s
8
10.
T = (3.6 div.)(2 ms/div.) = 7.2 ms
1
1
= 138.89 Hz
prf = =
T 7.2 ms
tp
1.6 div.
Duty cycle =  100% =
 100% = 44.4%
T
3.6 div.
11.
a.
T = (9  1)μs = 8 μs
c.
prf =
d.
Vav = (Duty cycle)(Peak value) + (1  Duty cycle)(Vb)
tp
2 s
Duty cycle =  100% 
 100% = 25%
T
8 s
Vav = (0.25)(6 mV) + (1  0.25)(2 mV)
= 1.5 mV  1.5 mV = 0 V
or
(2  s)(6 mV)  (2  s)(6 mV)
Vav =
=0V
8s
e.
Veff =
12.
b.
tp = (3  1)μs = 2 μs
1
1

= 125 kHz
T 8 s
(36  106 )(2 s)  (4 106 )(6 s)
= 3.46 mV
8 s
Eq. 24.5 cannot be applied due to tilt in the waveform.
(Method of Section 13.6)
Between 2 and 3.6 ms
1
(3.4 ms  2 ms)(2 V) + (3.6 ms  3.4 ms)(7.5 V) + (3.6 ms  3.4 ms)(0.5 V)
2
Vav =
3.6 ms  2 ms
1
(1.4 ms)(2 V) + (0.2 ms)(7.5 V) + (0.2 ms)(0.5 V)
2
=
1.6 ms
2.8 V + 1.5 V + 0.05 V
=
= 2.719 V
1.6
CHAPTER 24
335
13.
Ignoring tilt and using 20 mV level to define tp
tp = (2.8 div.  1.2 div.)(2 ms/div.) = 3.2 ms
T = (at 10 mV level) = (4.6 div.  1 div.)(2 ms/div.) = 7.2 ms
3.2 ms
t
 100% = 44.4%
Duty cycle = p  100% 
T
7.2 ms
Vav = (Duty cycle)(peak value) + (1  Duty cycle)(Vb)
= (0.444)(30 mV) + (1  0.444)(10 mV)
= 13.320 mV + 5.560 mV
= 18.88 mV
14.
Vav = (Duty cycle)(Peak value) + (1  Duty cycle)(Vb)
tp
(decimal form)
Duty cycle =
T
(8  1)  s
=
= 0.35
20  s
Vav = (0.35)(20 mV) + (1  0.35)(0)
= 7 mV + 0
= 7 mV
15.
Using methods of Section 13.8:
A1 = b1h1 = [(0.2 div.)(50 μs/div.)][(2 div.)(0.2 V/div.)] = 4 μsV
A2 = b2h2 = [(0.2 div.)(50 μs/div.)][(2.2 div.)(0.2 V/div.)] = 4.4 μsV
A3 = b3h3 = [(0.2 div.)(50 μs/div.)][(1.4 div.)(0.2 V/div.)] = 2.8 μsV
A4 = b4h4 = [(0.2 div.)(50 μs/div.)][(1 div.)(0.2 V/div.)] = 2.0 μsV
A5 = b5h5 = [(0.2 div.)(50 μs/div.)][(0.4 div.)(0.2 V/div.)] = 0.8 μsV
Vav =
16.
Using the defined polarity of Fig. 24.57 for υC, Vi = 5 V, Vf = +20 V
and τ = RC = (10 k)(0.02 μF) = 0.2 ms
a.
336
(4  4.4  2.8  2.0  0.8)  sV
= 117 mV
120  s
υC = Vi + (Vf  Vi)(1  et/τ)
= 5 + (20  (5))(1  et/0.2 ms)
= 5 + 25(1  et/0.2 ms)
= 5 + 25  25et/0.2 ms
υC = 20 V  25 Vet/0.2 ms
CHAPTER 24
b.
c.
Ii = 0
iC =
E  C 20 V   20 V  25 V e
=
R
10 k 
 t / 0.2 ms

 = 2.5 mAet/0.2 ms
d.
17.
υC = Vi + (Vf  Vi)(1  et/RC)
= 8 + (4  8)(1  et/20 ms)
= 8  4(1  et/20 ms)
= 8  4 + 4et/20 ms
= 4 + 4et/20 ms
υC = 4 V(1 + et/20 ms)
τ = RC = (2 k)(10 μF)
= 20 ms
18.
Vi = 10 V, Vf = 2 V, τ = RC = (1 k)(1000 μF) = 1 s
υC = Vi + (Vf  Vi)(1  et/τ)
= 10 V + (2 V  10 V)(1  et)
= 10  8(1  et)
= 10  8 + 8et
υC = 2 V+ 8 Vet
19.
Vi = 10 V, Ii = 0 A
CHAPTER 24
Using the defined direction of iC
(10 V  2 V) t/τ
e
iC =
1k
τ = RC = (1 k)(1000 μF) = 1 s
8 V t
iC = 
e
1k
and iC = 8mAet
337
20.
τ = RC = (5 k)(0.04 μF) = 0.2 ms (throughout)
υC = E(1  et/τ) = 20 V(1  et/0.2 ms)
(Starting at t = 0 for each plot)
a.
T=
1
1
=
= 2 ms
f 500 Hz
T
= 1 ms
2
5τ = 1 ms =
b.
T=
T
2
1
1
=
= 10 ms
f 100 Hz
T
= 5 ms
2
5τ = 1 ms =
c.
T=
1T 
 
5 2 
1
1
=
= 0.2 ms
f 5 Hz
T
= 0.1 ms
2
T 
5τ = 1 ms = 10  
2
21.
The mathematical expression for iC is the same for each frequency!
τ = RC = (5 k)(0.04 μF) = 0.2 ms
20 V  t / 0.2 ms
= 4 mAet/0.2 ms
and iC =
e
5 k
a.
T=
1
T
= 2 ms, = 1 ms
500 Hz
2
5τ = 5(0.2 ms) = 1 ms =
338
b.
T=
c.
T=
T
2
1
T
= 10 ms,
= 5 ms
100 Hz
2
1T 
5τ = 1 ms =  
5 2 
1
T
= 0.2 ms,
= 0.1 ms
5000 Hz
2
T 
5τ = 1 ms = 10  
2
CHAPTER 24
22.
τ = 0.2 ms as above
T=
1
= 2 ms
500 Hz
5τ = 1 ms =
0
T
2
T
: υC = 20 V(1  et/0.2 ms)
2
T
 T: Vi = 20 V, Vf = 20 V
2
υC = Vi + (Vf  Vi)(1  et/τ)
= 20 + (20  20)(1  et/0.2 ms)
= 20  40(1  et/0.2 ms)
= 20  40 + 40et/0.2 ms
υC = 20 V+ 40 Vet/0.2 ms
T
23.
3
T : Vi = 20 V, Vf = +20 V
2
υC = Vi + (Vf  Vi)(1  et/τ)
= 20 + (20  (20))(1  et/τ)
= 20 + 40(1  et/τ)
= 20 + 40  40et/τ
υC = 20 V  40 Vet/0.2 ms
υC = Vi + (Vf  Vi)(1  et/RC)
Vi = 20 V, Vf = 20 V
υC = 20 + (20  20)(1  et/RC)
T
= 20 V (for 0  )
2
For
T
 T, υi = 0 V and υC = 20 Vet/τ
2
τ = RC = 0.2 ms
with
T
T
= 1 ms and 5τ =
2
2
3
T , υi = 20 V
2
υC = 20 V(1  et/τ)
For T 
For
3
T  2T, υi = 0 V
2
υC = 20 Vet/τ
CHAPTER 24
339
24.
τ = RC = 0.2 ms
T
2
Vi = 10 V, Vf = +20 V
T
0 :
2
υC = Vi + (Vf  Vi)(1  et/τ)
= 10 + (20  (10))(1  et/τ)
= 10 + 30(1  et/τ)
= 10 + 30  30et/τ
υC = +20 V  30 Vet/0.2 ms
5τ = 1 ms =
25.
T
 T:
2
Vi = 20 V, Vf = 0 V
υC = 20 Vet/0.2 ms
1
1
=
= 5.31 M
2 fC 2 (10 kHz)(3 pF)
(9 M  0)(5.31 M    90)
Zp =
= 4.573 M 59.5
9 M   j 5.31 M 
Zp:
XC =
Zs:
CT = 18 pF + 9 pF = 27 pF
1
1
= 0.589 M
XC =
=
2 fC T 2 (10 kHz)(27 pF)
Zs =
Vscope =
(1 M  0)(0.589 M    90)
= 0.507 M 59.5
1 M   j 0.589 M 
(0.507 M    59.5)(100 V 0)
ZsVi =
Z s  Z p (0.257 M   j 0.437 M ) + (2.324 M   j 3.939 M )
50.7 106 V   59.5
1
= 10 V 0 =
(100 V 0)
6
10
5.07  10   59.5
 Zs =  Z p = 59.5
=
26.
Zp: XC =
1
1
= 3.333 M
=
5
C (10 rad/s)(3 pF)
(9 M  0)(3.333 M )
= 3.126 M 69.68
9 M   j 3.333 M 
1
1
XC =
= 0.370 M
=
5
C (10 rad/s)(27 pF)
Zp =
Zs:
(1 M  0)(0.370 M    90)
= 0.347 M 69.68
1 M   j 0.370 M 
  Z p = Z s
Zs =
340
CHAPTER 24
Vscope =
(0.347 M    69.68)(100 V 0)
ZsVi =
Z s  Z p (0.121 M   j 0.325 M ) + (1.086 M   j 2.931 M )
34.70  106 V   69.68
3.470  106   69.68
1
 10 V 0 =
(100 V 0)
10
=
CHAPTER 24
341
Chapter 25
1.
2.
I:
a.
no
b.
no
c.
yes
d.
no
e.
yes
II:
a.
yes
b.
yes
c.
yes
d.
yes
e.
no
III:
a.
yes
b.
yes
c.
no
d.
yes
e.
yes
IV:
a.
no
b.
no
c.
yes
d.
yes
e.
yes
b.
i=
c.
2Im  2
2
2

 1 + cos(2ωt  90)  cos(4ωt  90) + cos(6ωt  90) +  
  3
15
35

Im 2Im   
=
1

  4 
2
2
2
  2

i = 2 I m 1   cos(2ωt  90) 
cos(4ωt  90) 
cos(6ωt  90)  
15
35
  4 3

2Im

d.
2
2
2
 

cos (2ωt  90) 
cos (4ωt  90) +
cos (6ωt  90) + 
i = 2 I m 1  +
15
35
  4 3

3.
342
a.
υ = 4 + 2 sin α
b.
υ = (sin α)2
CHAPTER 25
c.
4.
i = 2  2 cos α
a.
b.
CHAPTER 25
343
5.
a.
b.
c.
344
CHAPTER 25
6.
a.
Vav = 100 V
Veff =
b.
7.
(100 V)2 +
(50 V)2 + (25 V) 2
= 107.53 V
2
Iav = 3 A
(2 A)2 + (0.8 A)2
= 3.36 A
2
Ieff =
(3 A)2 +
a.
Veff =
(20 V) 2 + (15 V)2 + (10 V)2
= 19.04 V
2
b.
Ieff =
(6 A) 2 + (2 A) 2 + (1 A)2
= 4.53 A
2
8.
PT = V0I0 + V1I1 cos θ1 +  + VnIn cos θn
(50 V)(2 A)
(25 V)(0.8 A)
cos 53 +
cos 70
= (100 V)(3 A) +
2
2
= 300 + (50)(0.6018) + (10)(0.3420)
= 333.52 W
9.
P=
10.
a.
DC: E = 18 V, Io =
b.
Ieff =
(20 V)(6 A)
(15 V)(2 A)
(10 V)(1 A)
cos 20 +
cos 30 +
cos 60
2
2
2
= 60(0.9397) + 15(0.866) + 5(0.5)
= 71.87 W
E 18 V
= 1.5 A
=
R 12 
ω = 400 rad/s:
XL = ωL = (400 rad/s)(0.02 H) = 8 
Z = 12  + j8  = 14.42  33.69
E
30 V/ 2 0
2.08 A
I= =
=
33.69
Z 14.42  33.69
2
 2.08 
i = 1.5 + 2 
 sin(400t  33.69)
 2 
i = 1.5 + 2.08 sin(400t  33.69)
CHAPTER 25
(1.5 A)2 +
(2.08 A) 2
= 2.10 A
2
345
c.
 2.08 A

  33.69  (12  0)
DC: υR = E = 18 V, VR = 
2


24.96 V
=
  33.69
2
 24.96 
υR = 18 + 2 
 sin(400t  33.69)
 2 
υR = 18 + 24.96 sin(400t  33.69)
d.
2
V Reff = (18 V ) +
e.
DC: VL = 0 V
(24.96 V ) 2
= 25.21 V
2
ω = 400 rad/s: VL =  2.08 A   33.69  (8  90)
2


16.64 V
=
56.31
2
υL = 0 + 16.64 sin(400t + 56.31)
11.
346
(16.64 V)2
= 11.77 V
2
f.
2
V Leff = 0 +
g.
2
P = I eff
R = (2.101 A)2 12  = 52.97 W
a.
DC: IDC =
b.
Ieff =
c.
υR = iR = i(12 )
= 24 + 24.96 sin(400t  33.69) + 6 sin(800t  53.13)
d.
Veff =
24 V
=2A
12 
ω = 400 rad/s:
Z = 12  + j(400 rad/s)(0.02 H) = 12  + j8  = 14.422  33.69
30 V 0
= 2.08 A 33.69 (peak values)
I=
14.422  33.69
ω = 800 rad/s:
Z = 12 Ω + j(800 rad/s)(0.02 H) = 12  + j16  = 20  53.13
10 V 0
= 0.5 A 53.13 (peak values)
I=
20  53.13
i = 2 + 2.08 sin(400t  33.69) + 0.5 sin(800t  53.13)
(2 A) 2 +
(2.08 A)2 + (0.5 A 2 )
= 2.51 A
2
(24 V) 2 +
(24.96 V) 2 + (6 V)2
= 30.09 V
2
CHAPTER 25
e.
DC: VL = 0 V
ω = 400 rad/s: VL = (2.08 A 33.69)(8  90)
= 16.64 V 56.31
ω = 800 rad/s: VL = (0.5 A 53.13)(16  90)
= 8 V 36.87
υL = 0 + 16.64 sin(400t + 56.31) + 8 sin(800t + 36.87)
12.
(0) 2 +
(16.64 V) 2 + (8 V) 2
= 13.06 V
2
f.
Veff =
g.
2
R = (2.508 A)2 12  = 75.48 W
PT = I eff
a.
DC: I = 
b.
Ieff =
c.
60 V
= 5A
12 
ω = 300 rad/s: XL = ωL = (300 rad/s)(0.02 H) = 6 
Z = 12  + j6  = 13.42  26.57
E = (0.707)(20 V) 0 = 14.14 V 0
E
14.14 V 0
= 1.054 A 26.57
I= =
Z 13.42  26.57
ω = 600 rad/s: XL = ωL = (600 rad/s)(0.02 H) = 12 
Z = 12  + j12  = 16.97  45
E = (0.707)(10 V) 0 = 7.07 V 0
7.07 V 0
E
= 0.417 A 45
I= =
16.97  45
Z
i = 5 + (1.414)(1.054)sin(300t  26.57)  (1.414)(0.417)sin(600t  45)
i = 5 + 1.49 sin(300t  26.57)  0.59 sin(600t  45)
(5 A) 2 +
(1.49 A) 2 + (0.59 A) 2
= 5.13 A
2
DC: V = IR = (5 A)(12 ) = 60 V
ω = 300 rad/s: VR = (1.054 A 26.57)(12  0)
= 12.648 V 26.57
ω = 600 rad/s: VR = (0.417 A 45)(12  0)
= 5 V 45
υR = 60 + (1.414)(12.648)sin(300t  26.57)  (1.414)(5)sin(600t  45)
υR = 60 + 17.88 sin(300t  26.57)  7.07 sin(600t  45)
d.
2
V Reff = (60 V ) +
CHAPTER 25
(17.88 V ) 2 + (7.07 V ) 2
= 61.52 V
2
347
e.
DC: VL = 0 V
ω = 300 rad/s:
ω = 600 rad/s:
VL = (1.054 A 26.57)(6  90) = 6.324 V 63.43
VL = (0.417 A 45)(12  90) = 5 V 45
υL = 0 + (1.414)(6.324)sin(300t + 63.43)  (1.414)(5)sin(600t + 45)
υL = 8.94 sin(300t + 63.43)  7.07 sin(600t + 45)
13.
f.
(8.94 V ) 2 + (3.54 V ) 2
= 6.8 V
V Leff =
2
g.
2
P = I eff
R = (5.13 A)2 12  = 315.8 W
a.
DC: I = 0 A
1
1
=
= 20 
ωC (400 rad/s)(125  F)
Z = 15   j20  = 25  53.13
E = (0.707)(30 V) 0 = 21.21 V 0
E
21.21 V 0
= 0.848 A 53.13
I= =
Z 25    53.13
i = 0 + (1.414)(0.848)sin(400t + 53.13)
i = 1.2 sin(400t + 53.13)
ω = 400 rad/s:
b.
c.
XC =
(1.2 A) 2
= 0.85 A as above
2
Ieff =
DC: VR = 0 V
ω = 400 rad/s: VR = (0.848 A 53.13)(15  0) = 12.72 V 53.13
υR = 0 + (1.414)(12.72)sin(400t + 53.13)
υR = 18 sin(400t + 53.13)
348
(18 V ) 2
= 12.73 V
2
d.
V Reff =
e.
DC: VC = 18 V
ω = 400 rad/s: VC = (0.848 A 53.13)(20  90)
= 16.96 V 36.87
υC = 18 + (1.414)(16.96)sin(400t  36.87)
υC = 18 + 23.98 sin(400t  36.87)
f.
2
V C eff = (18 V ) +
g.
2
P = I eff
R = (0.848 A)2 15  = 10.79 W
(23.98 V ) 2
= 24.73 V
2
CHAPTER 25
14.
a.
400
400
cos 4ωt
cos 2ωt 

3
15 
= 63.69 + 42.46 sin(2ωt + 90)  8.49 sin(4ωt + 90)
ω = 377 rad/s:
e = 63.69 + 42.46 sin(754t + 90)  8.49 sin(1508t + 90)
e=
200

DC: XL = 0  VL = 0 V
1
ω = 754 rad/s: XC = 1 =
= 1330 
ωC (754 rad/s)(1  F)
XL = ωL = (754 rad/s)(0.1 H) = 75.4 
Z = (1 k 0)  75.4  90 = 75.19  85.69
E = (0.707)(42.46 V) 90 = 30.02 V 90
Z(E)
(75.19  85.69)(30.02 V 90)
= 1.799 V 94.57
Vo =
=
Z + ZC 75.19  85.69 + 1330    90
ω = 1508 rad/s:
XC =
1
1
=
= 6631.13 
ωC (1508 rad/s)(1  F)
XL = ωL = (1508 rad/s)(0.1 H) = 150.8 
Z = (1 k 0)  150.8  90 = 149.12  81.42
E = (0.707)(8.49 V) 90 = 6 V 90
Z(E)
(149.12  81.42)(6 V 90)
Vo =
=

Z + ZC 149.12  81.42 + 6631.13    90
= 1.73 V 101.1
υo = 0 + 1.414(1.799)sin(754t  94.57)  1.414(1.73)sin(1508t  101.1)
υo = 2.54 sin(754t  94.57)  2.45 sin(1508t  101.1)
15.
b.
Voeff =
c.
P=
(2.54 V ) 2 + (2.45 V ) 2
= 2.50 V
2
( V eff ) 2 (2.50 V ) 2
=
= 6.25 mW
R
1k
i = 0.318Im + 0.500 Im sin ωt  0.212Im cos 2ωt  0.0424Im cos 4ωt +  (Im = 10 mA)
i = 3.18  103 + 5  103 sin ωt  2.12  103 sin(2ωt + 90)
 0.424  103 sin(4ωt + 90) + 
i  3.18  103 + 5  103 sin ωt  2.12  103 sin(2ωt + 90)
DC: Io = 0 A, Vo = 0 V
ω = 377 rad/s;
XL = ωL = (377 rad/s)(1.2 mH) = 0.452 
1
1
=
= 13.26 
XC =
C  377 rad/s  (200  F)
Z = 200   j13.26  = 200.44  3.79
I = (0.707)(5  103)A 0 = 3.54 mA 0
ZLI
(0.452  90)(3.54 mA 0)
= 7.98 μA 93.66
Io =
=

ZL + Z
j 0.452  + 200   j13.26 
CHAPTER 25
349
Vo = (7.98 μA 93.66)(200  0) = 1.596 mV 93.66
ω = 754 rad/s: XL = ωL = (754 rad/s)(1.2 mH) = 0.905 
XC =
1
1
=
= 6.63 
ωC (754 rad/s)(200  F)
Z = 200   j6.63  = 200.11  1.9
I = (0.707)(2.12 mA) 90 = 1.5 mA 90
Z LI
(0.905  90)(1.5 mA 90)
= 6.8 μA 181.64
Io =
=
Z L + Z
j 0.905  + 200   j 6.63 
Vo = (6.8 μA 181.64)(200  0) = 1.36 mA 181.64
υo = 0 + (1.414)(1.596  103)sin(377t + 93.66)
 (1.414)(1.360  103)sin(754t + 181.64)
3
υo = 2.26  10 sin(377t + 93.66) + 1.92  103 sin(754t + 1.64)
16.
a.
b.
17.
350
60 + 70 sin ωt + 20 sin(2ωt + 90) + 10 sin(3ωt + 60)
+20 + 30 sin ωt  20 sin(2ωt + 90) + 5 sin(3ωt + 90)
DC: 60 + 20 = 80
ω: 70 + 30 = 100  100 sin ωt
2ω: 0
3ω: 10 60 + 590 = 5 + j8.66 + j5 = 5 + j13.66 = 14.55 69.9
Sum = 80 + 100 sin ωt + 14.55 sin(3ωt + 69.9)
20 + 60 sin α + 10 sin(2α  180) + 5 sin(3α + 180)
0
 4 sin(3α  30)
5 + 10 sin α +
DC: 20  5 = 15
α:
60 + 10 = 70  70 sin α
2α: 10 sin(2α  180)
3α: 5 180  4 30 = 5  [3.46  j2] = 8.46 + j2
= 8.69 166.7
Sum = 15 + 70 sin α + 10 sin(2α  180) + 8.69 sin(3α + 166.7)
iT = i1 + i2
= 10 + 30 sin 20t
 0.5 sin(40t+ 90)
+20 + 4 sin(20t + 90) + 0.5 sin(40t+ 30)
DC: 10 A + 20 A = 30 A
ω = 20 rad/s: 30 A 0 + 4 A 90 = 30 A + j4 A = 30.27 A 7.59
ω = 40 rad/s: 0.5 A 90 + 0.5 A 30
= j0.5 A + 0.433 A + j0.25 A
= 0.433 A  j0.25 A = 0.5 A 30
iT = 30 + 30.27 sin(20t + 7.59) + 0.5 sin(40t  30)
CHAPTER 25
18.
e = υ1 + υ2
= 20  200 sin 600t + 100 sin(1200t + 90) + 75 sin 1800t
+ 50 sin(1800t + 60)
10 + 150 sin(600t + 30)
+0
DC: 20 V  10 V = 10 V
ω: 600 rad/s: 200 V 0 + 150 V 30 = 102.66 V 133.07
ω = 1200 rad/s: 100 sin(1200t + 90)
ω = 1800 rad/s: 75 V 0 + 50 V 60 = 108.97 V 23.41
e = 10 + 102.66 sin(600t + 133.07) + 100 sin(1200t + 90) + 108.97 sin(1800t + 23.41)
CHAPTER 25
351
Chapter 26
1.
2.
3.
4.
5.
Zi =
Ei
V
1.05  1.00 V 50 mV

; Ii = R 
= 1.064 mA
Ii
R
47 
47 
E
1.05 V
= 986.84 Ω
Zi = i 
I i 1.064 mA
Ei
120 V 0
= 19.35 Ω  10.8° = 19 Ω + j3.623 Ω

I i 6.2 A   10.8
XL
3.623 
f = 60 Hz: R = 19 Ω, L =
= 9.61 mH

2 f 2 (60 Hz)
Zi =
Ei1

20 mV
= 10 μA
2 k
a.
I i1 
b.
Z i2 
c.
Ei3  I i3 Z i3 = (1.5 mA)(4.6 kΩ) = 6.9 V
Z i1
Ei2
I i2
E g  Eo

1.8 V
= 4.5 kΩ
0.4 mA
4 V  3.8 V 0.2 V
= 0.1 mA(p  p)

2 k
2 k
Rs
E
3.8 V(p  p )
Zo = o 
= 38 kΩ
I o 0.1 mA( p  p )
Io =

Eopeak  Eg peak  VRpeak = 2 V  0°  40 × 103 V  0° = 1.96 V  0°
VRpeak
40 mV
= 43.96 µA
Rs
0.91 k
E 1.96 V 0
Zo = o 
= 44.59 kΩ
43.96  A
IR
Ipeak =
6.

Eopeak  2 0.6 V(rms) = 0.849 V


Eo( p p )  2 Eo( peak ) = 2(0.849 V) = 1.697 V
E g  Eo
18 V  1.697 V
= 51.5 µA(p  p)
2 k
Rs
E 1.697 V( p  p)
Zo = o 
= 32.95 kΩ
I o 51.5  A(p  p )
Io =
352

CHAPTER 26
7.
Zo =
Eo p p
I o p p

Eg p p  VRp p
I o p p

0.8 V  0.4 V
= 10 kΩ
40  A
VR p p  2 div  0.2 V/div. = 0.4 V
E g p p  4 div  0.2 V/div. = 0.8 V
I o p p 
8.
VR p p
10 k

0.4 V
= 40 μA
10 k
Ei = IiZi = (10 µA  0°)(1.8 kΩ  0°) = 18 mV  0°
E i(peak)  2 (18 mV) = 25.46 mV
Ei( p p ) = 2(25.46 mV) = 50.92 mV
A NL 
9.
a.
b.
Eo
4.05 V180
= 79.54  180° = 79.54

Ei 50.92 mV 0
Eo
RL
(5.6 k)
= 392.98
 A NL
 (3200)
Ei
5.6 k  40 k
RL  Ro
E
E E
AT  o  o  i
E g Ei E g
A 
with Ei =
AT 
10.
Zi E g
Zi  Rg
and
Ei
Zi

E g Zi + Rg
Eo
Zi
(2.2 k)

 (392.98)
= 320.21
Ei Zi  Rg
2.2 k  0.5 k
1400 mV
= 1200
1.2 mV 0
192 mV
= 160
A 
1.2 mV
 A

Ro  RL  NL  1
 A

A NL 

 1200 
 4.7 k 
 1
 160

 30.55 k
11.
a.
Eo
RL
 A NL
Ei
RL  Ro
2 k
160 = A NL
= A NL (0.0667)
2 k  28 k
A NL = 2398.8
A 
CHAPTER 26
353
b.
c.
Eo = IoRL = (4 mA)(2 kΩ) = 8 V
E
A = o = 160
Ei
Eo
8 V

= 50 mV
Ei =
160 160 V
Ii =
Zi =
12.
E g  Ei
Rg

70 mV  50 mV
= 50 µA
0.4 k
Ei 50 mV
= 1 kΩ

I i 50  A
Ri
RL  Ro
(3200)(2.2 k)
=
5.6 k  40 k
= 154.39
a.
Ai =  A NL
b.
 Rg  Z i 
A iT   AT 
 RL 
 A Z   Rg  Z i 
   i 

 Z i  R g   RL 
A iT   A

Zi
RL  Z i
   A NL
RL
RL  Ro  RL

  A NL
Zi
RL  Ro
 (3200)(2.2 k)
5.6 k  40 k
 154.39

13.
c.
Same result since Ii = Ig
a.
A G  A2
Ri
RL
 ( 392.98) 2
A = A NL
2.2 k
5.6 k
 6.067 ×104
A G   A A i
 (392.98)(154.39)
 6.067 ×104
RL
RL  Ro
5.6 k


 (3200) 
 5.6 k  40 k 
 392.98
A i =  A NL
Ri
RL  Ro
2.2 k


 (3200) 

 5.6 k  40 k 
 154.39
354
CHAPTER 26
b.
AT  A
Zi
2.2 k


 (392.98) 
= 320.21
 2.2 k  0.5 k 
Zi  Rg
A iT   AT
Rg  Z i
RL
 0.5 k  2.2 k 
 (320.21) 
 = 154.39

5.6 k
 Rg  Ri 
 0.5 k  2.2 k 
 ( 320.21) 2 
= 4.94 × 104
A GT   A2T 




5.6 k
 RL 
A GT   AT A iT  (320.21)(154.39) = 494 × 104
14.
a.
Ai 
Io
Z
  A i
Ii
RL
(160)(0.75 k)
2 k
 60

b.
A GT 
 Rg  Ri 
PL
 A2T 
Pg
 RL 
AT  A
Zi
Z i  Rg
(160)(0.75 k)
 104.35
0.75 k  0.4 k
 0.4 k  0.75 k 
A GT  (104.35) 2 


2 k
= 6.261 × 103

15.
16.
a.
AT  A1  A2 = (30)(50) = 1500
b.
A iT  AT
c.
A i1   A1
 1 k 
  (30) 
= 15
 2 k 
RL1
A i2   A2
 2 k 
 (50) 
= 12.5
 8 k 
RL2
ZiL
RL
 1 k 
= 187.5
 (1500) 
 8 k 
Zi1
Zi2
d.
A iT  A i1  A i2 = (15)(12.5) = 187.5 as above
a.
AT  A1  A2  A3
 
6912 = (12) A 2 (32)
A2 = 18
CHAPTER 26
355
b.
A i1 
4=
 A1 Zi1
RL1

 A1 Zi1
Zi2
 (12)(1 k)
Z i2
Z i2 = 3 kΩ
c.
A i3 
 A3 Zi3
RL3

 (32)(2 k)
2.2 k
= 29.09
A iT  A i1  A i2  A i3
= (4)(26)(29.09)
= 3.025 × 103
17.
a.
z11 =
E1
 Z1  (Z 2  Z3 )
I1 I  0
2
z11 =
Z1Z 2  Z1Z 3
Z1 + Z 2  Z3
Z3I 2
Z1 + Z 2  Z3
(Z3I 2 )(Z1 )
E1 = I1Z1 =
Z1 + Z 2  Z3
I=
z12 =
Z1I 3
E1

I 2 I  0 Z1 + Z 2  Z3
1
z21 =
E2
Mirror image of z12
I1 I  0
2
 z 21 
z22 =
Z1Z 3
Z1 + Z 2  Z3
E2
Mirror image of z11
I 2 I 0
1
 z 22  Z3  (Z1  Z 2 )

356
Z1Z3  Z 2 Z3
Z1 + Z 2  Z3
CHAPTER 26
18.
b.

a.
z11 
E1
I1 I  0
2
z11  R4  R2  ( R1  R3 )
`
 R4 
z12 
R2 ( R1  R3 )
R1  R2  R3
E1
I 2 I 0
1
R2 (I 2 )
( R1  R2 )  R3
E1  I R2  I 2 R4
I 

and z12 =
R2 R3I 2
 R4 I 2
R1  R2  R3
E1
R2 R3
R R  R4 ( R1  R2  R3 )

 R4  2 3
R1  R2  R3
I 2 R1  R2  R3
E2 = I R3  I1 R4
R2 (I1 )
CDR: I  =
( R1  R3 )  R2
R2 R3I1
E2 =
+ I1R4
R1  R2  R3
and z21 =
z 22 
E2
R2 R3
R R  R4 ( R1  R2  R3 )

 R4  2 3
R1  R2  R3
I1 R1  R2  R3
E2
I 2 I 0
1
Z22 = R4  R3  ( R1  R2 )
R ( R  R2 )
= R4  3 1
R3  ( R1  R2 )
CHAPTER 26
357
19.
a.
y11 =
I1
E1 E  0
2
YT  Y1  (Y2  Y3 )

Y1 (Y2  Y3 )
Y1  Y2  Y3

Y1Y2  Y1Y3
Y1  Y2  Y3
Nodal Analysis:
V [Y1  Y2  Y3 ]  E2Y2
V  = I1/Y1
and
 I1
[Y1  Y2  Y3 ]  E 2 Y2
Y1
y12 =
I1
 Y1Y2

E 2 E  0 Y1  Y2  Y3
1
y21 =
I2
Mirror image of y12
E1 E  0
2
 y 21 
y22 =
 Y1Y2
Y1  Y2  Y3
I2
Mirror image of y11
E2 E 0
1
y 22  YT  Y2  (Y1  Y3 )

20.
a.
y11 =
Y1Y2  Y1Y3
Y1  Y2  Y3
I1
E1 E  0
2
y11  Y1  Y2  Y4
 Y1 

358
Y2 Y4
Y2  Y4
Y1 (Y2 + Y4 )  Y2 Y4
Y2  Y4
CHAPTER 26
I2
(using the above diagram)
E1 E  0
y21 =
2
I
 1
I2
I 
1 
 (E  + E) =   2  2    I 2 

Y1
 Y2 Y4 
 Y2 Y4 
E1 =
 Y  Y2 
I
YY
and E1 = I 2  4
with y21 = 2   2 4

E1
Y2  Y4
 Y4 Y2 
y11 =
I2
E1 E  0
2
I
 1
I1
I 
1 
 (E  + E  ) =   1  1   I1 

Y3
 Y2 Y4 
 Y2 Y4 
E2 =
and y12 = 
Y2 Y4
= y21
Y2  Y4
I2
E2 E 0
y22 =
y22 = Y3 + Y2  Y4 = Y3 +
1
=
Y2 Y4
Y2  Y4
Y3 (Y2 + Y4 ) + Y2 Y4
Y2  Y4
21.
h11 =
E1
ZZ
= ZT = Z1  Z 2  1 2
Z1 + Z 2
I1 E  0
2
Using the above figure:
Z1 (I1 )
CDR: I2 =
Z1 + Z 2
h11 =
I2
Z1
=
I1 E  0 Z1 + Z 2
2
CHAPTER 26
359
h12 =
E1
E2 I 0
1
VDR: E1 =
h12 =
Z1E 2
Z1 + Z 2
E1
Z1
=
E 2 I  0 Z1 + Z 2
1
Using above figure:
h22 =
I2
Z Z  Z 2 Z3
: Z  = Z 3  (Z1  Z 2 )  1 3
Z1 + Z 2  Z3
E2 I 0
1
1 Z1  Z 2 + Z3
h22 =

Z  Z1Z3  Z 2 Z3
22.
a.
h11 =
E1
I1 E  0
2
= Z i  R1  (R 2  R 3  R 4 )
h12 =
E1
E2 I 0
1
E1 = I R2  I 2 R4
R3 (I 2 )
I 
R1  R2  R3
R2 R3I 2
 E1 
 I 2 R4
R1  R2  R3
E
E2
and I2 = 2 
Z  R4  R3  ( R1  R 2 )





R2 R3
E2

E1 = 
 R4  
 R1  R2  R3
  R  R3 R1  R2 R3 
 4 R1  R2  R3 
and h12 =
360
E1
R2 R3  R4 ( R1  R2  R3 )

E 2 R1 R3  R2 R3  R4 ( R1  R2  R3 )
CHAPTER 26
Z  R2  R3  R4
(Z  )(I1 )
I R1 
Z  + R1
RI
I  1 1
R1  Z 
I R3 

R4 I 
R4  R1 (I1 ) 



R4  R3 R4  R3  R1  Z  
R1 R4 I1
( R3  R4 )( R1  Z )
I 2 =  I R1  I R3 
 Z I1
R1 R4 I1

Z  + R1 ( R3  R4 )( R1  Z )
 Z

R1 R4
I2



I1 E  0  Z  + R1 ( R3  R4 )( R1  Z ) 
h22 =
2

R1 R4 
1 
Z 

R1  Z  
R3  R4 
I2
1

E 2 I  0 ZT
h22 =
1
ZT  R4  R3  ( R1  R2 )
1
h22 =
R4  R3  ( R1  R2 )
A Y- conversion would have simplified the problem to on similar to Fig 26.70.
23.
h11 =
E1
I1 E  0
2
Y   Y1  (Y2  Y3 )
Y1Y2  Y1Y3
Y1  Y2  Y3
1 Y1  Y2 + Y3
h11 =

Y  Y1Y2  Y1Y3
Y 
h21 =
I2
I1 E  0
2
CHAPTER 26
361
From above figure:
 Z3I1
I1/Y3
CDR: I2 =

Z3 + Z 2 1/Y3  1/Y2
1/Y3
I2
 Y2


I1 E  0 1/Y3  1/Y2 Y2  Y3
and h21 =
2
h12 =
E1
E2 I 0
1
Z 3E2
1/Y3E 2

Z3 + Z 2 1/Y3  1/Y2
Y2 E 2
And E1 =
Y2  Y3
VDR: E1 =
E1
Y2
=
E 2 I  0 Y2 + Y3
with h12 =
1
h22 =
I2
Y2  Y3
Y =
(from above figure)
E2 I 0
Y2 + Y3
1
h22 =
I2
Y2 Y3
= Y =
E2 I 0
Y2 + Y3
1
24.
h11 =
E1
1
1


I1 E  0 YT Y1  Y2  Y4
2
h21 =
I2
 Z1 (I1 )
1/ Y1 (I1 )
: CDR  I2 =

I1 E  0
Z1 + Z 2 + Z 4 1/Y1  1/Y2  1/Y4
2
=
and h21 = 
362
1/ Y1 (I1 )
Y2 Y4  Y1Y4  Y1Y2
Y1Y2 Y4
Y2 Y4
Y2 Y4  Y1Y4  Y1Y2
CHAPTER 26
h11 =
E1
E2 I 0
1
VDR: E1 

1/Y1 (E 2 )
1/Y1  1/Y2  1/Y4
Y2 Y4
Y2 Y4  Y1Y4  Y1Y2
and h12 =
h22 =
Z1 (E 2 )
Z1 + Z 2 + Z 4
I2
= YT (using the above figure)
E2 I 0
1
YT = Y3  Y1  Y2  Y4
Y1Y2 Y4
= Y3 
Y1Y2  Y1Y4  Y2 Y4
25.
a.
b.
Eq. 26.45:
hf
Ai =

1  ho Z L

a.
= 47.62
Zi =
h f ZL
hi (1  ho Z L )  hr h f Z L
50(2 k)
 99
1 k(1  0.05)  (4  104 )(50)(2 k)
hr h f Z L
E1
 hi 
I1
1  ho Z L
 1 k 
b.
 1 
1 
(2 k)
 40 k 
Eq. 26.46:
A 
26.
50
Zo =
ho 
CHAPTER 26
(4  104 )(50)(2 k)
 961.9 
 1 

1 
(2
k
)
 40 k 
1
hr h f
hi  Rs

1
 200 k
1
(4  104 )(50)

40 k
1 k  0
363
27.
Z11 = 1 kΩ  0°, z12 = 5 kΩ  90°, z21 = 10 kΩ  0°, z22 = 2 kΩ  j4 kΩ, ZL = 1 kΩ  0°
z z
E
(5 k 90)(10 k)
= 9,219.5 Ω  139.40°
 z11  12 21  1 k 
I
z 22  Z L
2 k  j 4 k  1 k
E
z z
(5 k 90)(10 k)
Zo = 2  z 22  12 21  2 k  j 4 k 
= 29.07 kΩ  86.05°
I2
Rs  z11
1 k  1 k
Zi =
28.
1/ y 22 Z L
1/ y 22 + Z L
ZL

1  y 22 Z L
1/ y 22  Z L 
 ZL 
E2 = y21E1 

1  y 22 Z L 

 ZL

I1 = E1y11 + y12E2 = E1y11 + y12   y 21E1 

 1  y 22 Z L  

I1
y y Z
 y11  12 21 L
E1
1  y 22 Z L
E
1
and Zi = 1 
y y Z
I1
y11  12 21 L
1  y 22 Z L
364
CHAPTER 26
1
Rs
 y12 E 2  y12 E 2  y12 Rs E 2


E1 =
1
Y
y11 Rs  1
y11 
Rs
Y   y11 
 y R E 
I2 = y21E1 + y22E2 = y22  12 s 2  + y22E2
 y11 Rs  1 
I2
y y R
  12 21 s + y22
E2
y11 Rs  1
and Zo =
E2
=
I 2 E 0
1
29.
30.
1
y y R
y 22  12 21 s
1  y11 Rs
z11z 22  z12 z 21 (4 k)(4 k)  (2 k)(3 k)
= 2.5 kΩ

4 k
z 22
z
2 k
h12 = 12 
= 0.5
z 22 4 k
z
3 k
= 0.75
h12 =  21 
z 22 4 k
1
1
h22 =
= 0.25 mS

z 22 4 k
h11 =
a.
h = h11h22  h12h21 = (103)(20 × 106)  (2 × 104)(100)
= 20 × 103  20 × 103 = 0

h
2  104
= 10 Ω
z11 = h = 0 Ω, z12 = 12 
Z 22
h12 20  106 S
h 21
100
1
z21 =
= 5 MΩ, z22 =
= 50 Ω

6
h 22
h 22
20  10 S
b.
y11 =
h12 2  104
1
1
= 103 S, y12 =
= 2 × 107 S

 3
h11
h11 10 
103 

h
100
y21 = 21  3
= 100 × 103 S, y22 = h = 0 S
h11
h11 10 
CHAPTER 26
365